1 

I 


J.  A.  HENDERSON'S  • 


JNTELLECTUAL  AND  ^PRACTICAL 


CORRECTED,  REVISED. 


AND  SUPPLEMENTED  BY  IMPORTANT  RULES  IN 


FRENCH,  SPMISH  AND  GERMAN 


SECOND     EDITION. 


IllFUj 

KANCISCO. 

•f873. 


OF  THE 

UNIVERSITY 


HENDERSON'S 


Intellectual  and  Practical 


LIGHTNING  CALCULATOR. 


BY 

J,  A.  HENDERSON, 

GRADUATE  OF  UNION  COLLEGE. 

.Author  of  Calculator,  Book  of  Blocks  Illus? 

trating  Roots,  and  the  New  Decirnal 

Method  of  Computing  Interest 

and  Imparting  the  same. 


ADDRESSED  TO. 

J.-.A.  HENDERSON^ 

SAN  FRANCISCO. 


ENTERED  according  to  Act  of  Congress,  in  tne  year  of  our  Lord  1872, 

By  J.  A.  HENDERSON, 
In  the  Office  of  Librarian  of  Congress,  Washington.. 


PREFACE. 


It  is  better  to  know  everything  about  something,  than 
something  about  everything. 

Early  ideas  are  not  usually  true  ideas,  but  need  to  be 
revised  and  re-revised.  Right  means  straight,  and  wrong 
means  crooked.  And  knowing  that  thought  kindles  at 
the  fire  of  thought,  we  do  not  hesitate  to  offer  any 
apology  for  presenting  to  the  public  some  new  seed- 
thoughts,  and  right  methods  of  operation  in  business 
calculations. 

The  practical  utility  of  this  book  consists  in  the  brevity, 
conciseness,  and  general  application  of  its  rules.  Par- 
ticular attention  is  invited  to  the  grand  improvements  in 
computing  time,  all  possible  cases  in  interest,  squaring 
and  multiplying  numbers,  dividing  and !  multiplying 
fractions,  an  infinite  number  of  ways  and  the  absolute 
right  method,  of  extracting  roots. 


TABLE  OF  CONTENTS. 


Lithograph  of  the  Author 

Title  Page 1 

Preface 

The  Arithmetical  Alphabet 5 

Numeration , 6 

Method  of  Acquiring  Multiplication  Table 7 

Method  of  Addition 8 

The  Lightning  Process  by  Combination 9 

Multiplication,  Useful  Contractions 10 

Bapid  Process  of  Marking  Goods . . : 12 

To  Multiply  and  Divide  by  the  Aliquot  parts  of  100  anc 

1,000 14 

Lightning  Process  of  Calculating  Interest 16 

Problems  in  Interest 20 

Method  of  Squaring  Numbers  by  their  Complement  and 

Supplement 27 

Method  of  Multiplying  Numbers 29 

Greatest  Common  Factor  or  Divisor 30 

Least  Common  Multiple 31 

Method  of  Adding  and  Substracting  Fractions 32 

General  Principles  of  Fractions 33 

Division  of  Fractions 35 

To  find  the  value  of  Currency  when  Gold  is  at  a  Stated 

Price 36 

Interest  Table  and  Form  for  making  Tables 37 

To  find  the  difference  of  time  between  two  dates,  and 

tell  the  day  of  the  week  from  the  day  of  the  month.  39 

Powers  and  Hoots 42 

Method  of  Extracting  Square  Root 44 

Infinite  number  of  ways  of  finding  the  Square  Boot  of 

any  number 45 

Absolute  right  method  of  extracting  Cube  Boot 47 

Bule  for  Extracting  Cube  Boot 49 

Examples  in  Cube  Boot . .  ^ 50 

Square  and  Cube  Boot  of  Fractions 52 

To  Find  the  Surface  of  Plane  Figures 54 

Method  of  Measuring  Land 55 

Method  of  Measuring  Grain 57 

Some  of  the  Miscellaneous  Weights  to  the  Bushel 58 

Short  Methods  in  Division  and  Multiplication 58 

Mental  Exercise  or  Mental  Calisthenics 62 

Squaring  and  Multiplying  Numbers 64 

Miscellaneous  Problems 66 

General  Information 72 


THE  LIGHTNING  CALCULATOR. 


The  arithmetical  alphabet,  as  written  and  read, 
g  g     M     rf  The  second 

§     1     g     1     §     |     §     I     g          is  two  times 
S     B     g  I    1    §  the  first;  the 

§  -5  3  £  <d  '£  S  "3  3  third  three 
"»t,  *,  f,  f,  f,  f,  1,  f,  ***<  times  the 
first,  etc.,  up  to  the  last.  All  numbers  larger 
than  nine  are  expressed  by  combining  two  or 
more  of  these  ten  letters  or  figures,  and  assign- 
ing different  values  to  them,  according  as  they 
occupy  different  places. 

Ten  is  expressed  by  combining  one  and  zero, 
thus,  10;  and  omitting  the  unity  or  denominator 
for  brevity;  two  and  zero  combined  make  twenty, 
thus,  20;  three  and  zero,  thus,  30,  etc.  A  hundred 
is  expressed  by  combining  the  one  and  two  zeros, 
thus,  100;  two  hundred,  thus,  200.  Ten  ones 
make  a  ten;  ten  tens  make  a  hundred;  ten  hun- 
dred make  one  thousand;  that  is,  mimbers  in- 
crease from  right  to  left  in  a  tenfold  ratio; 
hence  each  removal  of  a  figure  one  place  towards 
the  left  increases  its  value  ten  times. 


6  HENDERSON'S 

The  different  values  which  the  same  figures 
have  are  called  simple  and  local  values.  The 
simple  value  of  a  figure  is  the  value  which  it 
expresses  when  it  stands  alone,  or  in  the  right 
hand  place. 

The  local  value  of  a  figure  is  the  increased 
value  which  it  expresses  by  having  other  figures 
placed  on  its  right. 


NUMERATION. 

The  art  of  reading  numbers  when  expressed 
by  figures  is  called  numeration,  and  can  be 
easily  acquired  from  the  following  table: 

lit    •    „ . §  § 

|    |    i    »    g    g     g    §    g    |    rf  ,| 


§  i 


C85  678  398  74G  391  872  281  9G4  358  123  243  795  937  456  144 
XV  XIV  XIII  XII  XI  X  IX  VIII  VII  VI  V  IV  III  II  I 

We  have  here  fifteen  periods  of  three  figures 
each,  beginning  at  the  right  hand.  The  first 
period,  which  is  occupied  by  units,  tens,  hun- 
dreds, is  called  units  period;  the  second  is  occu- 
pied by  thousands,  tens  of  thousands,  hundreds 
of  thousands,  and  is  called  thousands  period; 
and  so  on,  the  orders  of  each  successive  period 
being  units,  tens  and  hundreds. 

The  figures  in  the  table  are  read  thus:  685 


LIGHTNING  CALCULATOR. 

tredeeillions,  678  duodeciHions,  398undecillions, 
746  decillions,  391  nonillions,  872  octillions, 
281  septillions,  964  sextillions,  358  quiiitillions, 
123  quadrillions,  243  trillions,  795  billions,  937 
millions,  456  thousands,  144  units  or  ones. 

To  read  numbers  expressed  by  figures:  Point 
them  off  into  periods  of  three  figures  each,  com- 
mencing at  the  right  liand;  then,  beginning  at  the 
left  hand,  read  the  figures  of  each  period  in  tlie 
same  manner  as  those  of  the  right  hana  figure  are 
read,  and  at  the  end  of  each  period  pronounce  its 
name. 

The  method  of  acquiring  the  multiplication 
table  is  of  great  importance,  and  is  represented 
thus: 

f  !    f  t  i  f  1  f  1 

Forms  the  first  line  of  the  multiplication  table, 
and  may  be  rehearsed  thus :  1  times  1  is  1 ;  2 
times  1  is  2;  3  times  1  is  3;  4  times  1  is  4;  5 
times  1  is  5;  6  times  1  is  6;  7  times  1  is  7;  8 
.times  1  is  8;  9  times  1  is  9. 

The  second  line  is  2  times  the  first,  thus: 

2  times  1  is  2. 

2  times  2  is  2  more  than  2,  or  4. 

2  times  3  is  2  more  than  4,  or  6. 

2  times  4  is  2  more  than  6,  or  8. 

2  times  5  is  2  more~than  8,  or  10. 

2  times  6  is  2  more  than  10,  or  12 

2  times  7  is  2  more  than  12,  or  14. 


8  HENDERSON'S 

2  times  8  is  2  less  than  18,  or  16. 

2  times  7  is  2  less  than  16,  or  14. 

2  times  6  is  2  less  than  14,  or  12. 

2  times  5  is  2  less  than  12,  or  10. 

2  times  4  is  2  less  than  10,  or  8. 

2  times  3  is  2  less  than  8,  or  6. 

2  times  2  is  2  less  than  6,  or  4. 

Thus  gaining  a  knowledge  of  addition  and 
subtraction,  and  in  fixing  in  the  understanding 
i  knowledge  of  the  table. 

The  third  line  is  three  times  the  first. 

The  fourth  line  is  four  times  the  first. 

The  fifth  line  i&  five  times  the  first. 

The  sixth  line  is  six  times  the  first. 

The  seventh  line  is  seven  times  the  first,  etc. 


ADDITION. 

3) 
2; 

5\ 

7;       Commence   at  the   units   column,   add 

6\  two  figures  at  once,    omitting  the  words 

4  and  and  are,   stopping  between  forty  and 

g)  fifty.     Thus:  10,  15,  32,  42,  writing  the  2 

3  at  the  right  of  the  6;  begin  again — 12,  17, 

2'  19,   writing  down  the   9;  carry  1  for  the 

4\  19  and  4  for  the  catch  figure,  making  59. 

59 


LIGHTNING   CALCULATOB. 
46 

53  \       Two  or  more  columns  may  be  added  in 

'47  3  a  similar  way 
98) 

76  EULE. — For  aaaing  one  or  more  columns, 
34)  commence  at  the  right  hand  column;  find  the 
,„  sum;  add  all  except  the  right  hand  figure  to 
gg'  the  second  column;  proceed  in  like  manner 
—  with  all  the  remaining  columns. 
519 


THE  LIGHTNING  PROCESS  BY  COMBINATION. 

First  four  rows  are  miscellaneous;  second 
four  are  the  complement  of  the  first,  taking  9 
as  the  base : 

76371236r< 

812367842 

176542051 

534256729 

236287632  - 

187632157 

823457948 

465743270 


4579831022 
18 

EULE. — Prefix  the  number  of  nines  to  the  odd 
row,  strike  a  line  and  subtract  the  number  of  nines. 


10  HENDERSON'S 

MULTIPLICATION: 

To  find  the  product  of  two  numbers,  when  the 
multiplicand  and  the  multiplier  each  contain^ 
but  two  figures. 

EXAMPLE  1. —  32 

21 

672 

EXPLANATION. — 1.  Multiply  the  units  of  the 
multiplicand  by  the  unit  figure  of  the  multiplier. 
Thus:  2x1  is  2:  Set  the  2  down;  multiply  the 
tens  in  the  multiplicand  by  the  unit  figure  in 
the  multiplier,  ana  the  units  in  the  multiplicand 
by  the  tens  figure  in  the  multiplier,  thus:  1x3 
is  3;  and  2x2  are  4;  add  these  two  products 
together,  3+4  are  7.  Set  down  the  7;  mul- 
tiply the  tens  in  the  multiplicand  by  the  tens  in 
the  multiplier,  thus  :  2x3  are  6,  the  whole 
amount,  672. 

USEFUL  CONTRACTIONS. 
To  multiply  two  figures  by  11. 
RULE, — Between  the  two  figures  write  their  sum,, 
thus: 

32 
11 

352 

Thus:  the  sum  of  3  and  2  are  5;  place  the  51 
between  the  3  and  2  for  the  product. 


CALCULATOR.  11 

34 
11 

374 

When  the  sum  of  two  figures  is  over  9,  in- 
crease the  left  hand  figure  by  1,, 

78 
11 

858 

Three  ones  multiplied  by  three  ones  are 
12321;  four  ones  by  four  ones  are  1234321; 
five  ones  by  five  ones  are  123454321,  etc.,  etc. 

To  multiply  any  number  of  nines  by  the  same 
number  of  nines,  thus:  9999999x9999999  are 
99999980000001.  Or,  the  square  of  any  number 
of  nines  is  as  many  nines  as  are  in  the  number 
minus  one,  eight  and  as  many  ciphers  as  nines, 
and  one. 

To  SQUABE  ANY  NUMBEB  ENDING  IN  FIVE. 

KULE. — Omit  the  five  and  multiply  the  number 
by  the  next  higher  number,  and  annex  tiventy^five 
to  the  product. 

What  is  the  square  85  ?    Ans.  7225. 
Explanation :  We  simply  omit  the  5,  and  mul- 
tiply the  8  by  9,  the  next  higher  number,  and 
annex  25. 

The  square  of  25  is  625. 
The  square  of  35  is  1225. 


12  HENDERSON'S 

The  square  of  45  is  2025 
The  square  of  65  is  4225 
The  square  of  75  is  5625  etc. 

For  multiplying  mixed  numbers :  2  J  by  2  J  is 
6£ — increase  2  by  1  and  multiply  by  2,  and  an- 
nex the  product  of  |  by  J,  increase  2  by  1,  since 
the  sum  of  the  fractional  parts  is  a  unit, 
21  by  25  is    6i 
*by2Jis    6i 
by  41  is  20J 
by  51  is  m 
by  61  is  425 
by  1?  is    24 
•y  2?  is    6f» 
<y  3?  is  12a         f 
,y  4?  is  20& 
>jy  5}  is  30,1 
,  by  61  is  42* 

RULE, — The  integer  increase  by  one;  multiply 
by  the  integer,  and  the  product  of  the  fractional 
parts  annex;  increase  the  integer  by  one  since  the 
mm  of the  fractional  parts  make  a  Knit, 

RAPID  PROCESS  OF  MARKING  GOODS, 

To  tell  what  an  article  should  retail  for  to 
make  a  profit  of  20  per  cent.,  is  done  by  re- 
moving the  decimal  point  one  place  to  the  left. 

For  instance,  if  hate  cost  $17.50  per  dozen* 
remove  the  decimal  point  one  place  to  the  left, 


LIGHTNING  CALCULATOR.  13 

making  $1.75 — what   they  should  be  sold  for 

apiece  to  gain  20  per  cent,  on  the  cost.  If  they 

cost  $13  per  dozen,  they  should  be  sold  for 
$1.30  apiece,  etc. 

RULE. — Remove  the  point  one  place  to  the  left, 
on  the  cost  per  dozen,  to  gain  20  per  cent.;  increase 
or  diminish  to  suit  the  required  rate. 

NOTE. — Remove  the  point  one  place  to  the  left,  for 
12  tens  make  120. 

TABLE 

For  marking  all  articles  bought  by  the  aozen. 

N.  B.  Most  of  these  are  used  in  business. 
To  make  20  per  cent.,  remove  the  point  one 
place  to  the  left. 

To  make  80  per  cent,  remove  tho  point  and  add  one  half  itself. 

«  60  "          "  "          "          "    third 

«  50  «          ««  ««          ««          «    fourth 

«  44  «i          «*  «          «  «    fifth 

"  40  « 

"  37H  *« 

«  35  4* 

«  33>i  <« 

30         « 
«         28         " 

«          25          " 
«<          12^      •« 

18^      «« 


seventh 

eighth 

ninth 

tenth 

twelfth  " 

fifteenth       itself. 

twentieth         " 

twenty -fourth4* 


subtract  ona  sixteeeuth. 


tweniy-pixth 


If  you  buy  one  dozen  shirts  for  $28,  what 
shall  I  retail  them  for  to  make  50  per  cent.? 
Ana.  $3.50. 


14  HENDERSON'S 

TABLE 

Of  the  Aliquot  parts  o/lOO  and  1000. 
N.  B.     Most  of  these  are  used  in  business. 

12*  is  X  part  of 100   8*  is  1-12  part  of 100 

25   is  2-8  or  X  of 100   16^  is  2-12  or  1-6  of 100 

37*  is  3-8  part  of  100   33  H  is  4-12  or  ^  of 100 

50   i*4-8or*of 100   66-£  is  8-12  or  %  of 100 

f»2*  is  &  part  of 100      83}$  is  10-12  or  5-6  of 100 

75      is  6-8  or  %  of 100    125      is  ^  part  of 1000 

87*  is  7-8  part  of 100  250      is  2-8  or  ^  of....  •....  1000 

6U  is  1-16  part  of 100    375      is  %  part  of 1000 

18?4  is  3-16  part  of 100    C25      is  %  part  of 1000 

3Ui  is  5-16  part  of 100    875      is  %  part  of 1000 

To  multiply  by  an  aliquot  part  of  100. 

KULE. — Take  such  a  part  of  the  multiplicand  as 
the  multiplier  is  part  of  100,  and  call  it  hundreds. 

To  multiply  by  an  aliquot  part  of  1000 :  Tako 
such  part  of  the  multiplicand  as  the  multiplier 
if  part  of  1000,  and  call  it  thousands. 
To  divide  by  the  aliquot  parts  of  100. 
To  divide  any  number  by  12 J:  remove  the 
pohit  two  places  to  the  left,   and  multiply  the 
quotient  by  8.     Multiply  the  quotient  by  8,  be- 
cause 12|  is  %  of  100. 

To   divide   any  number  by  25:  remove  the 
point  two  places  to  the  left,  and  multiply  by  4. 
345-^-25 
3.45 
4 

Ans.        13.80 


LIGHTNING  CALCULATOR.  15 

.   To  divide  any  number  by  50:   remove  the 
point  two  places  to  the  left  and  multiply  by  2. 

.75-50 

2 

1.50 

To  divide  any  number  by  75:  remove  tho 
point  two  places  to  the  left,  multiply  by  4  and 
divide  by  3;  because  75  is  f  of  100 . 

To  divide  by  the  aliquot  parts  of  1000. 

To  divide  any  number  by  125:  remove  the 
point  three  places  to  the  left  and  multiply  by  8. 
Kemove  the  point  fchree  places  to  the  loft  to 
divide  by  a  thousand  and  multiply  by  8;  because 
125  is  £  of  a  thousand. 

Thus:         34G7+425  9,712.29+425 

8  8 


Ans.        27.736  Ans.     77.69832 

Etc.,  for  all  other  examples. 

To  divide  any  number  by  250 :  remove  the 
point  three  places  to  the  left  and  multiply  by  4. 

4.357^-250 
4 

Ans.        17.428 

357.25-250 
4 

Ans.      1.42900 


16  HENDEESON'S 

HENDERSON'S  LIGHTNING   PROCESS  OF  CALCU- 
LATING INTEREST. 

The  base  of  our  system  of  notation  being  10, 
numbers  increase  and  diminish  in  a  tenfold 
ratio;  increasing  from  the  decimal  point  to  the 
left,  and  decreasing  from  the  decimal  point  to- 
wards the  right.  Hence,  to  divide  any  number 
by  10,  remove  the  point  one  place  to  the  left. 

To  divide  any  number  by  100,  remove  the 
point  two  places  to  the  left. 

To  divide  any  number  by  1000,  remove  the 
point  three  places  to  the  left. 

To  multiply  any  number  by  10,  remove  the 
point  one  place  to  the  right. 

To  multiply  any  number  by  100,  remove  the 
point  two  places  to  the  right. 

To  multiply  any  number  by  1000,  remove  the 
point  three  places  to  the  right. 


INTEREST. 

Since  the  interest  is  generally  a  part  of  the 
principal,  the  method  of  calculating  it,  will 
come  under  the  method  of  dividing.  The  rule 
establishes  the  time  when  a  dollar  makes  a 
cent,  and  we  remove  the  point  two  places  to 
the  left;  for  one  hundredth  of  the  principal 
equals  the  interest.  In  ten  times  that  time,  a 
dollar  makes  ten  cents,  and  we  remove  the  point 


LIGHTNING  CALCULATOB.  17 

one  place  to  the  left,  because  a  tenth  of  the 
principal  is  the  interest;  in  one  tenth  of  the 
same  time  a  dollar  makes  a  mill,  and  we  remove 
the  point  three  places  to  the  left,  because  one 
thousandth  of  the  principal  equals  the  interest. 

RULE. — The  reciprocal  of  the  rate,  or  the  rate  in- 
verted, indicates  the  time  when  the  decimal  point 
can  be  removed,  two  places  to  the  left  in  all  cases;  ten 
times  that  time  one  place  to  the  left,  and  one  tenth  of 
the  same  time  three  places  to  the  left.  Increase  or 
dimmish  the  results  to  suit  the  time  given. 

TKe  arithmetical  alphabet  is  \,  f ,  |,  \ ,  -f->  f > 
•J,  \,  f,  and  0;  -J-  inverted  is  1;  f  inverted  is  J; 
•f-  inverted  is  -J-,  etc.  If  the  rate  is  1  per  cent, 
per  month,  one  inverted  gives  the  time  when  a 
dollar  makes  a  cent,  and  the  point  removed  two 
places  to  left,  shows  the  interest  in  every  case. 

Ten  times  one  inverted,  or  ten  months,  a  dol- 
lar makes  ten  cents,  and  the  point  being  removed 
one  place  to  the  left,  all  examples  for  that  rate  and 
time  are  calculated.  One  tenth  of  one  month, 
or  three  days,  a  dollar  makes  a  mill,  and  the 
point  removed  three  places  to  the  left,  shows  the 
interest  in  all  examples  for  that  rate  and  time. 
We  remove  the  point  one  place  to  tha  left,  be- 
cause a  tenth  of  the  principal  is  the  interest. 
We  remove  the  point  two  places,  because  a  hun- 
dredth of  the  principal  is  the  interest.  We 
remove  the  point  three  places,  because  a  thou- 


18 


HENDERSON'S 


sandth  of  the  principal  is  the  interest.  To  reach 
all  other  time,  simply  increase  or  dimmish  the 
results  to  suit  the  time  given. 

$600.00  @  1  per  cent,  per  mo.  for  two  months, 
remove  the  point  two  places  to  the  left  =  $6. 00 
the  interest  for  one  month.  Twice  $6.00,  the 
interest  for  one  month,  is  $12.00,  the  interest 
for  two  months.  The  interest  on  the  same 
amount  for  three  days  is  .60  cts.,  simply  remov- 
ing the  point  three  places  to  the  left.  The  in- 
terest for  ten  months  on  the  same  amount,  would 
be  $60.00.  Simply  removing  the  point  one 
place  to  the  left,  $60.00,  the  interest  for  ten 
months,  plus  $12,00,  the  interest  for  two  months 
is  $72.00,  the  interest  for  one  year. 

By  this  method  we  can  calculate  an  infinite 
number  of  examples  in  a  moment  when  working 
from  the  base.  At  one  per  cent,  per  month: 


Thus. 


$2 

9, 
11, 
22, 


5J67.35 
861.50 
746.75 
463.25 
5J38.40 
0100.50 


LIGHTNING    CALCULATOR, 


19 


By  this  method  a  world  of  work  is  done  in  the 
twinkling  of  an  eye,  and  the  way  opened  to  the 
answer  of  every  example  in  interest. 

The  rate  is  2  per  cent,  per  month,  2  inverted 
is  -|,  or  15  clays,  the  point  removed  two  places  to 
the  left,  all  examples  are  calculated  for  that  rate 
and  date,  10  times  half  a  month  or  five  months^ 
the  point  removed  one  place  to  the  left,  all  ex- 
amples are  calculated.  One  tenth  of  15  days, 
or  a  day  and  a  half,  the  point  removed  three 
places  to  the  left,  all  examples  are  performed 
for  that  time 


*"•     £3 


$10, 

25, 
2, 
9, 


0:00.50 

650.00 
475.30 
460.50 

2150.31 


Simply  increasing  or  diminishing  the  results ; 
we  find  the  answer  for  any  other  time. 


20  *  HENDERSON'S 

PROBLEMS  IN  INTEREST. 

PROBLEM  1. — What  is  the  interest  of  $50  for 
4  years  at  6  per  cent. 

SOLUTION. — Bemoving  the  point  one  place  to 
the  left,  we  have  $5.00  the  interest  for  20 
months.  J?or  40  months,  it  is  $10.00;  8  months, 
being  the  fifth  of  40  months,  the  interest  would 
be  $2.00;  $10.00  plus  $2.00  is  $12.00  the  interest 
for  48  months,  or  4  years. 

PROBLEM  2. — What  is  the  interest  of  $10.00 
for  2  years,  at  5  per  cent  ?  Simply  remove  the 
point  one  place  to  left,  and  you  have  the  in- 
terest. 

PROBLEM  3. — What  is  the  interest  of  $48.00 
for  6  years,  at  5  per  cent.? 

PROBLEM  4. — What  is  the  interest  of  $70.00 
for  7  years,  at  5  per  cent.? 

PROBLEM  5.— What  is  the  interest  of  $68.00 
for  5  years,  at  6  per  cent.? 

PROBLEM  6. — What  is  the  interest  of  $70.00  for 
2  years,  at  5  per  cent.? 

PROBLEM  7. — What  is  the  interest  of  $75.00 
for  5  years,  at  3  per  cent.? 

PROBi.Ejyf  8,— What  is  the  interest  of  $120.00 
for  3  years,  at  5  per  cent.? 

PROBLEM  9.— What  is  the  interest  of  $100.00 
for  10  years,  at  6  per  cTent.? 

PROBLEM  10.— What  i$  tlie  interest  of  $140.00 
for  12  years,  at  5  per  p 


LIGHTNING  CALCULATOB.  21; 

PROBLEM  11.— What  is  the  interest  of  $150.00 
for  5  years,  at  3  per  cent.? 

PROBLEM  12.— What  is  the  interest  of  $145.00 
for  6  years,  at  5  per  cent.? 

PROBLEM  13.— What  is  the  interest  of  $200.00 
for  10  years,  at  8  per  cent.? 

PROBLEM  14. — What  is  the  interest  of  $250.00 
for  3  years,  at  8  per  cent.? 

PROBLEM  15.— What  is  the  interest  of  $500.00 
for  9  years,  at  8  per  cent.? 

PROBLEM  16.— What  is  the  interest  of  $50.00 
for  2  years  and  2  months,  at  2  per  cent.? 

PROBLEM  17.— What  is  the  interest  of  $80.00 
for  8  years  and  6  months,  at  6  per  cent.? 

PROBLEM  18.— What  is  the  interest  of  $90.00 
for  5  years  and  6  months,  at  6  per  cent.? 

SOLUTION. — Remove  the  point  one  place  to  the 
left,  we  have  $9.00  the  interest  for  20  months. 
The  interest  would  be  3  times  $9.00,  which  is 
$27.00  for  5  years'.  The  interest  for  6  months 
would  be  one  tenth  of  $27.00,  which  is  $2.70, 
which  added  to  $27.00,  makes  $29.70,  Ans. 

PROBLEM  19.— What  is  the  interest  of  $90.00 
for  12  years  and  10  months,  at  6  per  cent.? 

PROBLEM  20.— What  is  the  interest  of  $200.00 
for  4  years  and  8  months,  at  3  per  cent.? 

PROBLEM  21.— What  is  the  interest  of  $70.00 
for  8  years  and  4  months,  at  2  per  cent.? 

PROBLEM  22.— What  is  the  interest  of  $225.00 
for  52  days,  at  7  per  cent.?  Ans.  $2.25., 


22  HENDERSON  S 

PROBLEM  23. — What  is  the  interest  of  $500.00 
for  26  days,  at  7  per  cent.?  Ans.  $2.50. 

PROBLEM  24.— What  is  the  interest  of  $500.00 
for  2  years,  6  months,  and  15  days,  at  4  per 
cent.? 

SOLUTION. — Remove  the  point  one  place  to  the 
left  we  have  §50.00,  the  interest  for  2  years  and 
6  months.  Removing  the  point  two  places  to 
the  left,  we  have  $5.00  the  interest  for  3  months; 
15  days  being  one  sixth  of  three  months,  we 
have  83J  cts.  the  interest  for  15  days,  which 
added  to  $50.00  makes  $50.83J,  Ans. 

PROBLEM  25.— What  is  the  interest  of  $200.00 
for  5  years,  9  months  and  18  days,  at  5  percent.? 

SOLUTION. — Removing  the  point  one  place  to 
the  left  we  have  $20.00,  the  interest  for  2  years. 
The  interest  for  5  years  would  be  2J  times 
$20.00,  or  $50.00.  The  interest  for  1  year  is 
$10.00;  for  9  months  it  would  be  |  of  $10.00, 
which  is  $7.50.  Removing  the  point  2  places  to 
the  left,  we  have  $2.00,  the  interest  for  72  days, 
the  interest  for  18  days  would  I  e  the  fourth  of 
$2.00,  which  is  50  cents,  added  to  $57.50,  would 
be  $58.00,  Ans, 

PROBLEM  26.— What  is  the  ii  terest  of  $700.00 
for  1  year,  7  months,  and  18  days,  at  6  per  cent.? 

PROBLEM  27.— What  is  the  interest  of  $250.00 
for  3  months,  at  1  per  cent,  per  month? 

SOLUTION. — Remove  the  point  two  places  to 
left,  we  have  $2.50,  the  interest  for  one  month. 


LIGHTNING  CALCULATOB.  23 

The  interest  for  3  months  would  be  three  times 
$2.50,  which  is  $7.50,  Ans. 

PROBLEM  28.— "What  is  the  interest  of  $60.00 
for  6  years.  4  months,  and  24  days,  at  5  per 
cent.? 

PROBLEM  29.— What  is  the  interest  of  $40.00 
for  1  year,  at  1  per  cent,  per  month  ? 

PROBLEM  30.— What  is  the  interest  of  $950.25 
for  9  months,  at  1  per  cent,  per  month? 

PROBLEM  31.— What  is  the  interest  of  $55.00 
for  11  months,  at  1  per  cent,  per  month? 

PROBLEM  32.— What  is  the  interest  of  $200.00 
for  10  months,  at  1  per  cent,  per  month? 

PROBLEM  33.— What  is  the  interest  of  $144.50 
for  15  months,  at  1  per  cent,  per  month? 

PROBLEM  34. — What  is  the  interest  of  $60.00 
for  22  months,  al  1  per  cent,  per  month  ? 

PROBLEM  35.— What  is  Ihe  interest  of  $600.00 
for  18  days,  at  10  per  cent,  per  ennum? 

SOLUTION. — Eeinove  the  point  two  places  to 
the  left,  we  have  $6.00,  the  interest  for  36  days. 
The  interest  for  18  days  is  one-half  of  $6.00, 
which  is  $3.00,  Ans. 

PROBLEM  36.—  WTiat  is  the  interest  of  $250.25 
for  35  days,  at  10  per  cent,  per  annum  ? 

PROBLEM  37.— What  is  the  interest  of  $360.50 
for  1  year,  at  10  per  cent,  per  annum  ?  $36.05, 
Ans. 

PROBLEM  38.— What  is  the  interest  of  $200.00 
for  72  days,  at  10  per  cent,  per  annum? 


24  HENDERSON'S 

PROBLEM  39.— What  is  the  interest  of  $80.00 
for  one  year,  at  f  per  cent,  per  month  ?  Ans. 
$8.00. 

PROBLEM  40.— What  is  the  interest  of  $500.00 
for  2  years,  at  f  per  cent,  per  month  ? 

PROBLEM  41.— What  is  the  interest  of  $250.00 
for  3  years,  at  -|  per  cent,  per  month? 

NOTE. — Remove  the  point  one  place  to  the 
left,  because  a  tenth  of  the  principal  is  the  in- 
terest. Two  places,  because  a  hundredth  of  the 
principal  is  the  interest,  etc. 

PROBLEM  52.— What  is  the  interest  of  $250.00 
for  one  month,  at  1  per  cent,  per  month? 

SOLUTION. — At  1  per  cent,  per  month,  one 
one  hundredth  of  the  principal  is  the  interest, 
we  therefore  remove  the  point  two  places  to  the 
left.  Ans.  $2.50.  Removing  the  point  two 
places  to  the  left,  we  have  the  answer. 

PROBLEM  43.— What  is  the  interest  of  $250.50 
for  2  months,  at  1  per  cent,  per  month  ? 

SOLUTION.— Removing  the  point  two  places, 
we  get  the  interest  $2.505  for  1  month;  for  2 
months,  the  interest  would  be  twice  $2.505, 
which  would  be  $5.01. 

PROBLEM  44.— What  is  the  interest  of  $100.00 
for  15  days,  at  1  per  cent,  per  month? 

SOLUTION. — Removing  the  point  two  places  to 
the  left,  we  get  the  interest  $1.00,  for  1  month. 
The  interest  for  15  days  would  be  one  half  of 
$1.00,  or  50  cents. 


LIGHTNING  CALCULATOR.  25 

PROBLEM  45. — What  is  the  interest  of  $145.00 
for  3  days,  at  1  per  cent,  per  month  ? 

SOLUTION. — Eemove  the  point  three  places  to 
the  left,  and  we  have  $1.45,  Ans. 

PROBLEM  46.— What  is  She  interest  of  $2000.00 
for  9  days,  at  1  per  cent,  per  month  ? 

SOLUTION. — Remove  the  point  three  places, 
we  have  the  interest  $2.00  for  3  days;  for  9 
days,  $6.00,  Ans. 

PROBLEM  47.— What  is  the  interest  of  $250.25 
for  12  days,  at  1  per  cent,  per  month  ? 

PROBLEM  48.— What  is  the  interest%of  $270.00 
for  10  months,  at  1  per  cent,  per  month  ?  Ee- 
move the  point  1  place  to  the  left.  Ans.  $27.00. 

PROBLEM  49.— What  is  the  interest  of  $350.00 
for  1  year,  at  1  per  cent,  per  month  ?  Kemove 
the  point  one  place,  we  have  the  interest  $35.00 
for  10  months,  for  one  year,  one  fifth  more, 
$42.00,  Ans. 

PROBLEM  50.— What  is  the  interest  of  $250.00 
for  11  months  and  3  days,  at  1  per  cent,  per 
month?  Interest  10  months,  $25.00;  interest  1 
month,  $2.50;  interest  3  days,  25  cents,  equals 
$27.75,  Ans. 

PROBLEM  51. — What  is  the  interest  of  $2,500.00 
for  1  year,  at  1  per  cent,  per  month? 

PROBLEM  52.— What  is  the  interest  of  $125.00 
for  33  days,  at  1  per  cent,  per  month  ? 

PROBLEM  53.— What  is  the  interest  of  $260.00 
for  24  days,  at  1J  per  cent,  per  month? 


28  HENDERSON'S 

SOLUTION. — Remove  the  point  two  places  to 
the  left,  we  have  the  interest  $2.60,  Ans. 

PROBLEM  54.— What  is  the  interest  of  $360.00 
for  1  month,  at  1J  per  cent,  per  month  ? 

SOLUTION. — Remove  the  point  two  places  to 
lie  left,  we  have  $3.60,  the  interest  for  24  days; 
add  i,  .90,  we  have  $4.50,  Ans. 

PROBLEM  55.— What  is  the  interest  of  $800.50 
for  8  months,  at  1J  per  cent,  per  month  ? 

SOLUTION. — Remove  the  point  one  place  to  the 
left.  Ans.  $80.05. 

PROBLEM  56.— What  is  the  interest  of  $500.00 
for  1  year,  at  1 J  per  cent,  per  month  ? 

SOLUTION. — Remove  the  point  one  place  to 
the  left,  we  have  the  interest  $50.00  for  8  months. 
For  4  months,  the  interest  would  be  $25.00, 
added  to  $50.00,  equals  $75.00,  Ans. 

PROBLEM  57.— What  is  the  interest  of  $900.00 
for  4  months,  at  1J  per  cent,  per  month? 

SOLUTION. — Remove  the  point  one  place  to  the 
left,  we  have  $90.00,  the  interest  for  8  months; 
for  4  months,  the  interest  would  be  one  half  of 
$90.00,  or  $45.00,  Ans. 

Removing  the  point  one  place  to  the  left, 
gives  the  interest  of  any  sum  for  8  months,  at 
1J  per  cent.,  increase  or  diminish  the  result  to 
suit  the  time  given 


LIGHTNING  CALCULATOR.  27 

METHOD  OF  SQUARING  NUMBERS  BY  THEIR  COM- 
PLEMENT AND  SUPPLEMENT, 

The  complement  of  a  number  is  the  difference 
between  the  number  and  some  particular  num- 
ber above  it. 

The  supplement  of  a  number  is  the  difference 
of  a  number  and  some  number  below  it. 

(99)2  =  9801.  Take  the  complement  of  99 
from  it,  call  it  hundreds,  and  add  the  square  of 
the  complement. 

EXPLANATION. — Let  N  equal  99,  and  C  equals 
1.  Then  N  plus  0  =  100.  N— C  =  98.  Mul- 
tiplying the  two  equations  together,  we  have 
N2— C2  =  9800.  Add  C2  to  both  menbers  o 
the  equation,  and  we  have  N2  —  9801,  the  square 
of  99. 

(98)2  =  9604.  Now  2,  the  complement  of  98 
from  98  =  96;  call  it  hundreds,  and  add  the 
square  of  2,  and  we  have  9604,  the  square  of  98. 

(97)2  =  9409.  The  complement  3  from  97  = 
94;  call  it  hundreds,  and  add  the  square  of  3, 
and  we  have  the  square  of  97. 

(96)2  =  9216.  The  complement  of  96  is  4;  4 
from  96  =  92,  call  it  hundreds,  and  add  the 
square  of  4,  and  we  have  the  square  of  96. 

(95)2  =  9025.  The  completent  of  95  is  5; 
95  —  5  90,  call  it  hundreds,  and  add  the 
square  of  5,  and  we  have  the  square  of  95. 

(101)2  =  10201.     The  supplement  of  101  is  1 ; 


28  HENDEKSON'S 

1  added  to  101  is  102,  call  it  hundreds,  and  add 
the  square  of  1,  and  we  have  10201  the  isquare 
of  101. 

(102)2  —  104U4.  The  supplement  is  2,  added 
to  102  is  104T  call  it  hundreds,  and  add  the 
square  of  2,  and  we  have  10404  the  square  of 
103. 

RULE. — WJien  ubove  the  base,  add  the  supple- 
ment,  call  it  hundreds,  and  add  the  square  of  the 
supplement,  call  it  hundreds,  because  the  number 
ivhen  increased  by  the  supplement,  is  multiplied  by 
one  hundred  in  this  case,  when  below  t  substract  the 
complement. 

(103)2  =  10609.  The  supplement  3  added, 
call  it  hundreds,  and  add  T!he  square  of  3. 

(104)2  =  10816. 

(1001)2=1002001.  The  supplement  is  1  added 
to  1001  =  1002,  call  it  thousands,  and  add  the 
square  of  1,  and  it  equals  1002001.  (1002)2  = 
1004004.  (1003)'= 1006009.  (1004)2= 1008016. 

(999)2  =  998001.  The  complement  is  1  from 
999  equals  998,  call  it  thousands,  and  add  the 
square  of  1,  and  we  have  the  square  of  the  num- 
ber. (998)2=996004.  (997)2=  994009.  (996)2 
= 992016.  (995) : = 99002d.  (994)2 = 988036, 
etc. 

Take  any  number  that  is  easy  to  multiply  by 
for  the  base  10,  20,  30,  50,  80,  100,  1000,  etc. 

92  =  81.     The  complement  of  9  is  1,  1  from 


LIGHTNING  CALCULATOR.  2t> 

9  leaves  8,  call  it  tens  and  add  the  square  of  1, 
and  we  have  the  square  of  9. 

82  =64.  The  complement  of  8  is  2,  2  from 
8  leaves  6,  call  it  tens,  and  add  the  square  of  2, 
and  we  have  the  square  of  8. 

(II)2  =  121.  The  supplement  of  11  is  1,  1 
added  to  11  is  12,  call  it  tens  and  add  the  square 
of  1,  and  we  have  the  square  of  11. 

(12)2  =  144.  The  supplement  is  2,  2  added 
to  12  is  14,  call  it  tens  and  add  the  square  of  2, 
and  we  have  the  square  of  the  number. 

(13) 3  =  169.  The  supplement  of  13  is  3,  3 
added  is  16,  call  it  tens  and  add  the  square 
of  3,  and  we  have  the  square  of  the  number. 

(14)3  =  196.     (15)3  =  225. 

(19)2  =  361.  The  complement  is  1,  1  from  19 
leaves  18,  18  multiplied  by  20,  equals  360,  add 
the  square  of  1,  and  we  have  the  square  of  the 
number. 

(18)2  =  324.  (17)2=289.  (16)s=256.  (21)2 
=441.  (22)2—484.  (49)2=2401.  The  com- 
plement is  1,  1  from  49  is  48,  call  it  fifties,  and 
add  the  square  of  1,  and  we  have  2401,  Ans. 

(51)2=2601.     (52)2=2704.     (53)2=2809. 

To  multiply  numbers. 

KULE. —  T/ie  product  of  any  two  numbers  is  the 
square  of  their  mean,  diminished  by  the  square  of 
half  their  difference. 

19x21=399.  The  mean  is  20,  the  square  of 
20  is  400;  400— I2  is  399,  the  product  of  19  X 


30  HENDEESON'S 

21,  18  x  22.  The  mean  is  20,  the  square  of 
20  is  400,  22is  4;  4  from  400  leaves  396,  the 
product,  17x23=391.  The  square  of  3  is  9;  9 
from  400  leaves  391  the  product. 

16  x  24=384.  The  square  of  4  is  16.  16  from 
400  leaves  384  the  product. 

15  x  25  =375.  The  square  of  5  is  25.  25  from 
400  leaves  375  the  product. 

29x31=899.  The  mean  is  30.  The  square 
is  900,  minus  the  square  of  1  is  899  their  pro- 
duct. 

28  x  32 = 896.  The  square  of  the  mean  is  900, 
minus  the  square  of  2  is  896  the  product. 

27x33=891.  26x34=884.  25x35=875. 

39x41=1599.  38x42=1596.  37x43=1591. 

36x44=1584.  35x45=1575.  34x46=1564. 

49x51=2499  48x52=2496.  47x53=2491. 


GREATEST    COMMON   FACTOR 
OR   DIVISOR 

What  is  the  greatest  common  divisor  of  21  and 
77.  Separating  the  numbers  into  their  prime 
factors  we  have  21=7x3,  77=7  Xll,  hence  7  is 
the  greatest  common  factor  or  the  greater  com- 
mon divisor  of  the  two  numbers. 

RULE. — Separate  the  numbers  into  their  prime 
factors.  Tlie  product  of  all  the  factors  that  are 
common  will  be  the  greatest  common  divisor. 


LIGHTNING  CALCULATOB.  31 

What  is  the  greatest  divisor  of  25  and  60.  25 
—5  X  5,  60=5  X  3  X  2  X  2  ?  Hence  5  is  the  greatest 
common  divisor. 

What  is  the  greatest  common  divisor  of  5,  15 
and  20? 

What  is  the  greatest  common  divisor  of  36, 
18,  24 and  12.  36=6x6, 18=6x3,  24=6x4  12 
=6  X  2  ?  Hence  6  is  the  greatest  common  factor 
or  divisor. 

What  is  tne  greatest  common  divisor  of  135 
and  225? 

What  i*  the  greatest  common  divisor  of  4,  8, 
12,  16? 

What  is  the  greatest  common  divisor  of  25 
and  75? 

What  is  the  greatest  common  divisor  of  13 
^nd  65? 

What  is  the  greatest  common  divisor  of  14 
and  42? 


LEAST  COMMON  MULTIPLE. 

A  multiple  of  a  number  is  any  number  which 
contains  it  as  a  factor. 

A  common  multiple  of  two  or  more  numbers 
is  any  number  which  contains  them  all  as  factors. 

The  least  common  multiple  of  two  or  mo*e 
numbers  is  the  least  number  which  contains 
them  all  as  factors.  Hence  it  follows  a  multiple 


32  HENDERSON'S 

of  a  number  must  contain  all  the  prime  factors 
of  that  number. 

A  common  multiple  of  two  or  more  numbers 
must  contain  all  the  prime  factors  of  those  num- 
bers. 

The  least  common  multiple  of  two  or  more 
numbers  must  be  the  least  number  that  contains 
sill  the  prime  factors  of  those  numbers. 

KULE. —  Tlie  product  of  all  the  prime  factors  of 
that  number  having  the  greatest  number  of  prime 
factors,  and  those  prime  factors  of  the  other  num- 
bers not  found  in  the  factors  of  the  number  taken, 
will  be  the  least  common  multiple. 

What  is  the  least  common  multiple  of  12  and 
18?  12=2X2X3,  18=2X3X3.  The  least  com- 
mon multiple  is  2X2X3X3  or  36. 

What  is  the  least  common  multiple  of  4  and  6? 

What  is  the  least  common  multiple  of  18  and 
36? 

What  is  the  least  common  multiple  of  4,  6,  8 
and  10? 

WTiat  is  the  least  common  multiple  of  2,  4,  6, 
9  and  18? 

What  is  the  least  common  multiple  of  2,  3,  4, 
5  and  6  ? 

RULE  FOE  ADDING  AND  SUBTRACTING  FRACTIONS. 

First  make  the  fractions  similar  by  reducing 
them  to  the  same  denominator.  Add  the  numer- 


LIGHTNING  CALCULATOR.  33 

ators  and  place  the  sum  over  the  common  denomin- 
ator. In  subtraction  ivrite  the  difference  of  the 
numerators  over  the  common  denominator. 

What  is  the  sum  of  -\  and  |,  |— ^V,  -J=*V>  A 
+&=« ,  Ans.  |+l=l,  4+f=l*. 

What  is  the  sum  of  T5¥  and  i=|. 

What  is  the  sum  of  -fa  and  4=lf  • 

What  is  the  sum  of  f  and  1=1 

What  is  the  sum  of  f  and  J-=l 

What  is  the  sum  of  f  and  -1=1 

From  f  subtract  ^=TV 

From  |  subtract  |.  |=Jf ,  |=^,  Ji~A=8V 

From  A  take  f  |=^,  A— &=&• 

What  is  the  sum  of  3£,  2J,  4J,  5J= 15f . 

Add  the  fractions  and  whole  numbers  sepa- 
rately. 

What  is  the  sum  of  9J,  6J,  7|=23J. 

From  8t  take  3£,  J=f ,  f— i=J.  8—3=5;  5 
+i=5i. 

From  23f  take  9J.     f=f ,  J=f,  f-*=fc  25 


GENERAL   PRINCIPLES    OF 
FRACTIONS. 

Multiplying  the  numerator  multiplies  the  frac- 
tion. 

Dividing  the  numerator  divides  the  fraction. 

Multiplying  the  denominator  divides  the  frac- 
tion. 


34  HENDERSON'S 

Dividing  the  denominator  multiplies  the  frac-  1 
tion. 

Multiplying  both  terms  of  the  fraction  by  the  ' 
same  number  does  not  change  its  value. 

Fractions  are  called  similar  when  they  have  a 
common  denominator,  as  f  ,  |,  f  ,  ^. 

Dissimilar  fractions  are  fractions  which  are 
not  alike,  as  f  ,  f  ,  |,  5,  f  . 

The  numerators  of  similar  fractions  only  can 
be  added, 

The  common  denominator  is  written  under 
the  sum  or  difference  . 

Multiply  T*T  by  8=ff=an. 

Multiply  A  by  14=j£=5. 

Multiply  40  by  f  =5x5=25. 

Multiply  3J  by  6.  Multiply  the  whole  num- 
ber and  fraction  separately.  6xJ=3,  6  x  3=18  x 
3=21. 

Multiply  4J  by  8.  8xJ—  2f,  8x4=32+2f= 
34|. 

Multiply  7|-  by  9.    9xi=4J,  9x7=63-)-4i= 


Multiply  8J  by  12.    12xi=6,  12x8=96+6= 
102. 

Multiply  7J  by  7£.    7|  x  7J=52.  T96 
Multiply  1\  by  7J=56J. 
Multiply  8|  by  8J=72|. 
Multiply  9f  by  9f=9 


f  OF  TBEB 

f  TTNIVERSIT^ 
LIGHTNING  CALCULATOB.         '^ElllfM -- 

DIVISION   OF  FRACTIONS. 

RULE. — Reduce  mixed  numbers  to  improper  frac- 
tions, and  whole  numbers  to  the  form  of  fractions; 
multiply  the  dividend  by  the  divisor  inverted,  or 
multiply  both  numerator  and  denominator  by  the 
least  common  midtiple  of  the  denominators  of  the 
fractional  parts. 

$3i=sJ£=±2£.  Simply  multiplying  numerator 
and  denominator  by  2. 

Divide  5J  by  2J-.  Multiply  both  numerator 
and  denominator  by  6,  the  least  common  mul- 
tiple of  2  and  3. 

Divide  25  by  £=50. 

Divide  21  by  3J=fJ=6TV 

To  divide  any  number  by  3| ,  remove  the  point 
one  place  to  the  left  and  multiply  by  3. 

Divide  20  by  3J .  Remove  the  point  one  place 
we  have  2,  2x3=6  Ans. 

Divide  27  by  3J=8TV- 

To  divide  any  number  by  2 J,  remove  the  point 
one  place  to  the  left  and  multiply  by  4. 

Divide  20^  by  2J.  Remove  the  point  one 
place  to  the  left  and  multiply  by  4. 

Removing  the  point  one  place  to  the  left 
makes  2-^,  2^X^=84-  Ans. 

To  divide  any  number  by  1-J-,  remove  the 
point  one  place  to  the  left  and  multiply  by  9. 
Divide  11  by  1-^—9^. 

Divide  any  number  by  5.     Remove  the  point 


36  HENDERSON'S 

one  place  and  multiply  by  2.  Removing  the 
point  one  place  to  the  left  divides  the  number 
by  10.  In  dividing  by  10  we  divide  by  a  num- 
ber twice  too  large;  therefore  we  multiply  by  2 
for  the  correct  result. 

To  divide  any  number  by  12J,  remove  the 
point  two  places  to  the  left  and  multiply  by  8. 

Divide  125  by  12J. 

Divide  47  5  by  12J. 

Divide  96  by  12J. 

Divide  99  by  12J. 

To  divide  any  number  by  25,  remove  the 
point  two  places  to  the  left  and  multiply  by  4. 

To  divide  any  number  by  33J,  remove  the 
point  two  places  to  the  left  aud  multiply  by  3. 

To  divide  any  number  by  50,  remove  the 
point  two  places  to  the  left  and  multiply  by  2. 

To  divide  by  66f ,  remove  the  point  two  places 
to  the  left,  divide  bv  2  •end  multiply  by  3. 


TO  FIND  THE  VALUE  OF  CURRENCY  WHEN  GOLD 
IS  AT  A  STATED  PRICE, 

When  gold  is  111|,  what  is  the  value  of  $1.00 
currency?  "We  take  the  100,  the  number  of  cents 
in  a  dollar,  as  the  -numerator,  and  the  value  of 
the  gold  as  the  denominator.  Simplify  the 
fraction  by  multiplying  the  numerator  and  de- 


LIGHTNING  CALCULATOR.  37 

nominator  by  9  and  we  have  T°F  of  a  dollar  or 
90  cents;  the  value  of  the  currency. 

When  gold  ts  109^,  what  is  the  value  of  $1.00 
currency? 

100          900          450  ~  319 


109  :       982  :       491  — 

When  currency  is  worth  75  cents,  what  is  the 
value  of  gold? 

inn          4.4 

^L  =;  |,   |  of  100  cents  equals  $1.33}. 
75          o     o 

When  gold  is  worth  105J,  what  is  the  value 
of  $1.00  currency? 

100       _  200    _  *  94166 
105!  "  Z  2lT  211 

RULE.  —  We  take  100,  the  number  of  cents  in  a 
dollar,  for  the  numerator,  and  the  value  of  gold  or 
currency,  as  the  case  may  be,  for  the  denominator. 
Simplify  the  fraction  l>y  annexing  ciphers  to  the 
numerator  and  dividing  by  the  denominator. 

INTEREST  TABLE  AND  FORM  FOR  MAKING  TABLES. 

Ehe  following  Table  gives  the  Interest  on  any 
amount  at  7  per  cent.,  by  simply  removing  the 
Doint  to  right  or  left,  as  the  case  ma,v  require: 


38 


HENDERSON'S 


Number  of  Days. 

$100 

$90 

$80 

$70 

1 

.0192 

.01726 

.01534 

.01342 

2 

.0384 

.03452 

.03058 

.02685 

3   

.0575 

.05178 

.04603 

.04027 

4  

.0767 

.06904 

.06137 

.05370 

5 

.0959 

08630 

.07671 

.06712 

6  

.1151 

.10356 

.09205 

.08055 

7  

.1342 

.12082 

.10740 

.09897 

8 

1532 

13808 

12274 

.10740 

9 

.1726 

.15534 

.13808 

.12089 

90 

1  7260 

1  5342 

1  38082 

1  .  20822 

93  

1.7836 

1.60521 

1.42685 

1.24849 

100  

1.9178 

1  82603 

1.53425 

1.24247 

$60 

$50 

$40 

$30 

$20 

.01151 

.00950 

.00767 

.00575 

.00384 

.02301 

.01918 

.  01534 

.01151 

.00767 

.03452 

.02877 

.02301 

.01726 

.01151 

.04603 

.02836 

.03068 

.02301 

.01536 

.05753 

.04795 

.03836 

.02877 

.01918 

.06904 

.05753 

.04603 

.03452 

.02313 

.08055 

.06712 

.05370 

.04027 

.02685 

.09205 

.07671 

.06137 

.04603 

.03068 

1.0356 

.08630 

.06904 

.05178 

.03452 

1.03562 

.86301 

.69041 

.51781 

.34521 

1.07014 

.89178 

.71342 

.53508 

.35671 

1.15065 

.95890 

.76712 

.57534 

.48356 

LIGHTNING  CALCULATOR.  39 

To  FIND  THE  DIFFERENCE  OF  TIME  BETWEEN 
Two  DATES  BY  THE  FOLLOWING  TABLE  : 

RULE. — Opposite  the  day  of  the  month  is  written 
the  number  of  days  of  the  year  ivhich  have  expired. 
Subtract  this  number  from  the  whole  number  of 
days  that  have  expired  at  the  last  date. 

Thus :  What  is  the  time  from  the  first  day  of 
March  to  the  27th  day  of  September?  The  1st 
day  of  March  we  find  by  the  table  that  60  days 
of  the  year  are  gone.  The  27th  day  of  Septem- 
ber we  find  that  270  days  are  gone.  Hence  270 
days  minus  60  days  equals  210  days,  the  time 
between  the  two  dates. 

To  FIND  THE  DAY  OF  THE  WEEK  FKOM  THE  DAY 
OF  THE  MONTH  BY  THE  SAME  TABLE  : 

Cast  the  sevens  out  of  the  day  of  the  month, 
the  ratio  of  the  month,  the  ratio  of  the  year 
which  is  3,  and  the  year.  One  of  a  remainder 
will  be  the  first  day  of  the  week,  two,  second, 
etc.  0  the  last  day  of  the  week.  The  ratio  of 
the  month  is  found  above  its  name.  The  ratio 
of  every  month  except  January  and  February 
is  one  more  in  Leap  Years. 


40 


HENDEKSONS 


3 

January 

6 
I  ebruary 

6 
March 

•2 

April 

4 
May 

0 
June 

1    1 

1   32 

1   60 

1   91 

1  121 

1  152 

2   2 

2   33 

2   61 

2   92 

2  122 

2  153 

3   3 

3   34 

3   62 

3   93 

3  123 

3  154 

4   4 

4   35 

4   63 

4   94 

4  124 

4  155 

5   5 

5   36 

5   64 

5   95 

5  125 

5  156 

6   6 

6   37 

6   65 

6   96 

6  126 

6  157 

7   7 

7   38 

7   66 

7   97 

7  127 

7  158 

8   8 

8   39 

8   67 

8   98 

8  128 

8  159 

9   9 

9   40 

9   68 

9   09 

9  129 

9  160 

10   10 

10   41 

10   69 

10  100 

10  130 

10  161 

11   11 

11   42 

11   70 

11  101 

11  131 

11  162 

12   12 

12   43 

12   71 

12  102 

12  132 

12  163 

13   13 

13   44 

13   72 

13  103 

13  132 

13  164 

14   14 

14   45 

14   73 

14  104 

14  134 

14  165 

15   15 

15   46 

15   74 

15  105 

15  135 

15  166 

16   16 

16   47 

16   75 

16  106 

16  136 

16  167 

17   17 

17   48 

17   76 

17  107 

17  137 

17  168 

18   18 

18   49 

18   77 

18  108 

18  138 

18  169 

19   19 

19   50 

19   78 

19  109 

19  139 

19  170 

20   20 

20   51 

20   79 

20  110 

20  140 

20  171 

21   21 

21   52 

21   80 

21  111 

21  141 

21  172 

22   22 

22   53 

22   81 

22  112 

22  142 

22  173 

23   23 

23   54 

23   82 

23  113 

22  143 

23  174 

24   24 

24   55 

24   83 

24  114 

24  144 

24  175 

25   25 

25   56 

25   84 

25  115 

25  145 

25  176 

26   26 

26   57 

26   85 

26  116 

26  146 

26  177 

27   27 

27   58 

27   86 

27  117 

27  147 

27  178 

28   28 

28   59 

28   87 

28  118 

28  148 

28  179 

29   29 

V 

29   88 

29  119 

29  149 

29  180 

30   30 

30   89 

30  120 

30  150 

30  181 

31   31 

31   90 

31  151 

LIGHTNING  CALCULATOB. 


2 

July 

5 

August 

1 

September 

3 

October 

6 

November 

1 

December 

1  182 

1  213 

1  244 

1  274 

1  305 

1  335 

2  183 

2  214 

2  245 

2  275 

2  306 

2  336 

3  184 

3  215 

3  246 

3  276 

3  307 

3  337 

4  185 

4  216 

4  247 

4  277 

4  308 

4  338 

5  186 

5  217 

5  248 

5  278 

5  309 

5  339 

6  187 

6  218 

6  249 

6  279 

6  310 

6  340 

7  188 

7  219 

7  250 

7  280 

7  311 

7  341 

8  189 

8  220 

8  251 

8  281 

8  312 

8  342 

9  190 

9  221 

9  252 

9  282 

9  313 

9  343 

10  191 

10  222 

10  253 

10  283 

10  314 

10  344 

11  192 

11  223 

11  254 

11  284 

11  315 

11  345 

12  193 

12  224 

12  255 

12  285 

12  316 

12  346 

13  194 

13  225 

13  256 

13  286 

13  317 

13  347 

14  195 

14  226 

14  257 

14  287 

14  318 

14  348 

15  196 

15  227 

15  258 

15  288 

15  319 

15  349 

16  197 

16  228 

36  259 

16  289 

16  320 

16  350 

17  198 

17  229 

17  260 

17  290 

17  321 

17  351 

18  199 

18  230 

18  261 

18  291 

18  322 

18  352 

19  200 

19  231 

.  19  262 

19  292 

19  323 

19  353 

20  201 

20  232 

20  263 

20  293 

20  324 

20  354 

21  202 

21  233 

21  264 

21  294 

21  325 

21  355 

22  203 

22  234 

22  265 

22  295 

22  326 

22  356 

23  204 

23  235 

23  266 

23  296 

23  327 

23  357 

24  205 

24  236 

24  267 

24  297 

24  328 

24  358 

25  206 

25  237 

25  268 

25  298 

25  329 

25  359 

26  207 

26  238 

26  269 

26  299 

26  330 

26  360 

27  208 

27  239 

27  270 

27  300 

27  331 

27  361 

28  209 

28  240 

28  271 

28  301 

28  332 

28  362 

29  210 

29  241 

29  272 

29  302 

29  333 

29  363 

30  21 

30  242 

30  273 

30  303 

30  334 

30  364 

31  21 

31  243 

31  304 

31  365 

42  HENDEKSON'S 

POWERS  AND   ROOTS. 

The  product  of  a  number  taken  any  number 
of  times  as  a  factor,  is  called  a  power  of  the 
number. 

A  root  of  a  number  is  such  a  number  as  taken 
some  number  of  times  as  factor  will  produce  a 
given  number. 

If  the  root  is  taken  twice  as  a  factor  to  pro- 
duce the  number,  it  is  the  square  root.  If  three 
times,  the  cube  root.  If  four  times,  the  fourth 
root,  etc. 

ILLUSTRATION. — 5  is  the  square  root  of  25. 
The  cube  root  of  125.     The  fourth  root  of  625, 
because  (5)2=25,  (5)3=125,  (5)4=625. 
(1)»=1  (1)3=1 

(2)2=4  (2)3=8 

(3)2=9  (3)3=27 

(4)2=16  (4)3=64 

(5)2=25  (5)3=125 

(6)2=r36  (6)3=216 

(7)2=49  (7)3=343 

(8)2=64  (8)3=512 

(9)2=81  (9)3=729 

(10)2=100  (10)3=10QO 

We  observe  that  the  square  of  any  one  of  the 
digits  is  less  than  100.  And  the  cube  of  any 
one  of  the  digits  is  less  than  1000.  Hence  the 
square  root  of  two  figures  cannot  give  more  than 
one  figure. 


LIGHTNING  CALCULATOR.  43 

Hence  if  we  begin  at  the  right  of  any  number 
and  separate  it  into  periods  of  two  figures  each, 
the  number  of  periods  would  be  the  same  as  the 
number  of  figures  in  its  square  root. 

In  order  to  understand  the  method  of  extract- 
ing square  root,  it  is  necessary  to  consider  how 
the  square  of  a  number  consisting  of  two  parts 
is  formed  from  those  parts. 

To  do  this  let  a  represent  any  number  what- 
ever, ~b  represent  any  other  number,  then  will 
a  -f-  &  represent  the  sum  and  (a  +  6)2  the  square 
of  the  sum  of  any  two  numbers,  but  since  the 
square  of  any  two  terms  is  the  square  of  the 
first,  plus  two  times  the  first  into  the  second,  plus 
the  square  of  the  second :  we  have  (a  -f  6)2=:a2  -f 
2  a  6  +  &2. 

ILLUSTRATIONS.  —  23  here  a =20  and  &— 3. 
HencO  (a  -j-  Vf  will  equal  (20  -j-  3)2.  In  applying 
the  above  formally,  commence  at  the  units  in- 
stead of  the  tens  to  find  the  square  of  the  num- 
ber. Thus  32  is  9,  two  times  3  into  2  is  12. 
Write  down  the  2  and  carry  the  1  to  the  square 
of  the  first  term  2,  and  we  have  529,  the  square 
of  23  and  23  is  the  square  root  of  529. 

The  square  of  any  number  of  terms  is  the 
square  of  the  first,  plus  two  times  the  first  into 
the  second,  plus  the  square  of  the  second,  plus 
two  times  the  sum  of  the  first  two  into  the  third, 
plus  the  square  of  the  third,  plus  two  tjie  sum  of 
the  first  three  into  the  fourth,  plus  the  square  of 


44  HENDERSON'S 

the  fourth,  etc.  Note — In  applying  the  above 
formula  commence  at  the  units  to  square  num- 
bers. 

METHOD  OF  EXTRACTING  SQUARE  BOOT. 


2    1 


625.  This  number  'contains  two  periods; 
hence  there  are  two  figures  in  the  roots.  The 
greater  square  below  6,  the  first  or  left  hand 
period  is  4,  the  root  of  which  is  2;  and  since 
there  are  two  figures  in  the  root,  2  will  stand  in 
the  tens  place  and  equal  20.  Hence,  we  sub- 
tract the  square  of  20,  which  is  400,  from  625, 
and  we  have  225  remaining.  We  have  found  a 
square  20  feet  on  a  side.  Now,  in  order  to  pre- 
serve the  square,  we  make  the  addition  on  two 
adjacent  sides.  Hence,  we  double  20,  the 
length  of  one  side,  and  get  40,  the  trial  divisor; 
dividing  225  by  40,  we  get  the  width  of  the  ad- 
dition, 5  feet;  adding  5  feet  to  40  feet,  the  width 
of  the  little  square  in  the  corner,  we  get  45,  the 
true  divisor.  Multiplying  45  by  5,  we  get  225, 
the  surface  of  the  addition.  Hence,  25  is  the 
length  of  one  side  of  a  square  that  contains  625 
square  feet. 


LIGHTNIN  A  />LCULATOB.  45 

ICO        J5625(100+20+5- 
1st  trial  divisor    200        10000 

1st  true       "         220          5625 
2d  trial       "         240          4400 

1225 
2dtrue       "          245          1225 

We  may  have  an  infinite  number  of  ways  for 
finding  the  square  root  of  any  number. 

Thus :  Presume  the  root  of  the  number  to  be 
divided  into  a  certain  number  of  equal  parts* 
Let  5  a  equal  the  square  root  of  15625.  Since 
the  square  of  the  root  is  equal  to  the  number 
(5a)2  or,  25a2=15625  and  a2=625,  and  a=25. 
5a  is  the  root  =125.  Presume  the  root  of  15625 
to  be  75a,  then  (25a)2  =625(a)2  =15625,  (a)2 
=  25,  a=5,  25a=125.  In  the  same  way,  we 
may  presume  the  root  to  be  divided  into  2,  3,  4, 
5,  or  any  number  of  equal  parts.  Hence  the 
rule :  Divide  any  number  by  the  square  of  two, 
extract  the  square  root  of  the  quotient,  "end  we 
have  one  half  of  the  root  of  the  number. 

Divide  any  number  by  the  square  of  three,  ex- 
tract the  square  root  of  the  quotient,  and  we  have 
one  third  of  the  root  of  the  number. 

Divide  any  number  by  the  square  of  four,  ex~ 
tract  the  square  root  of  the  quotient,  and  we  have 
a  fourth  of  the  square  root  of  the  number,  etc. 


45  HENDERSON'? 

What  is  the  square  root  of  9604? 
What  is  the  square  root  of  2401  ? 
What  is  the  square  root  of  225? 
What  is  the  square  root  of  64? 


CUBE    HOOT. 

RELATION  OF  CUBE  TO  BOOT. 

I3  =    1 

03  _       Q       Bv  observation  we   see    that  the 

zj      O  ^ 

33  —  27  entire  part  of  the  cube  root  of  any 
43  =  64  number  below  1000  will  be  less  than 
53  =125  jo  and  will,  therefore,  contain  but 

/•»3    Q-|  /•» 

73  Hqlq   one  figure-     The  entire   part  of  the 

gs  —512    cu^e    ro°t  °f    a  number  containing 

93  =739   four,  five  or  six  figures,  will  contain 

103=1000   two    figures,    and    so    on  with    the 

larger  numbers. 

Hence:  If  we  begin  at  the  right  of  any  num- 
ber, and  separate  it  into  periods  of  three  figures 
each,  the  number  of  periods  will  equal  the 
number  of  figures  in  the  entire  part  of  the  cube 
root.  The  cube  of  the  highest  denomination 
will  be  found  in  the  left  hand  period.  The 
cube  of  the  two  highest  will  be  found  in  the  two 
left  hand  periods,  etc. 

A  cube  of  any  number  of  terms,  is  the  cube  of 
the  first  term,  plus  three  times  the  square  of  the 
first  into  the  second,  plus  three  times  the  first 
into  the  square  of  the  second,  plus  the  cube 


LIGHTNING  CALCULATOE.  47 

of  the  second,  plus  three  times  the  square  of 
the  sum  of  the  first  two  into  the  third,  plus  three 
times  the  sum  of  the  first  two  into  the  square 
of  the  third,  plus  the  cube  of  the  third,  etc. 

METHOD  OF  EXTRACTING  CUBE  BOOT. 

10000  31 29531125(100+20+5 

1  000  000 

30000  Trial  divisor. 
6000  953 125 

4001  728000 


36400  1st  true  divisor  225  125 

6000  225 125 

800  


43200  2d  trial  divisor. 
1800 
25 

45025  2d  true  divisor. 

EXPLANATION. — We  separate  the  number  into 
periods  of  three  figures  each,  by  placing  small 
digits  over  the  periods.  "We  find  the  greatest 
cube  in  the  first  or  left  hand  period,  which  is  1, 
the  cube  root  of  which  is  1;  and  since  there  are 
three  periods,  fhere  will  be  three  figures  in  the 
root,  and  this  1  will  stand  in  hundreds  place 
and  equal  100.  We  will  presume  the  linear 
edge  of  a  cubical  block  to  be  100  feet.  The 
surface  of  one  side  will  be  100  times  100,  or 


48  HENDERSON'S 

10000  square  feet,  and  the  solid  contents  will 
be  100  times  10000,  or  1000000  solid  feet;  sub- 
tracting this  number  from  the  given  number, 
we  have  953125  feet  remaining. 

To  increase  this  cube  and  preserve  the  cubical 
form,  we  must  make  the  addition  on  three  ad- 
jacent sides;  and  since  10000  is  the  surface  of 
one  side,  three  times  10000,  or  30000  will  be 
the  surface  of  three  sides,  which  forms  the  trial 
divisor;  dividing  the  dividend  by  this  number, 
we  find  20,  the  thickness  of  the  addition;  but 
besides  these  three  large  square  pieces,  there 
are  three  parallelopipedons,  the  length  of  each 
100  feet,  the  width  20  feet.  Hence,  these  sur- 
faces would  be  20  times  300,  or  6000.  The  little 
cube  is  20  feet  each  way,  the  surface  of  one  side 
of  it  would  be  20  times  20,  or  400,  adding  30000, 
6000  and  400,  we  have  36400,  the  sum  of  the 
surfaces  of  one  side  of  each  of  the  pieces  making 
the  addition.  Multiply  36400  by  20,  the  thick- 
ness of  the  addition,  we  have  728000  the  solid 
content  of  the  addition.  Subtracting  from  the 
last  dividend,  we  have  225125  feet  to  be  still 
added.  The  next  trial  divisor  is  three  sides  of 
the  complete  cube;  by  observation  we  see  that 
36400  lacks  of  being  three  sides  of  the  complete 
'  cube.  One  side  of  each  of  the  parallelopipedons 
and  two  sides  of  the  little  cube.  Hence,  by 
bringing  down  6000  and  doubling  400,  adding 
the  36400,  we  obtain  three  sides  of  the  com- 


LIGHTNING  CALCULATOR.  49 

plete  cube,  or  tlie  trial  divisor;  dividing,  we 
find  the  thickness  to  be  5  feet.  The  three  de- 
ficiencies, the  length  of  one  is  120  feet,  the 
length  of  three  would  be  360  feet,  the  width  5 
feet,  the  sum  of  the  surfaces  of  one  side  of  each 
would  be  five  times  360  feet,  which  is  1800; 
the  surface  of  one  side  of  the  small  cube  would 
be  5  times  5,  or  25,  adding  to  the  43200,  we 
have  45025  feet;  multiply  by  5  to  get;  the  solid 
contents  of  the  last  addition;  if  there  were 
another  figure  in  the  root,  we  would  simply 
bring  down  the  1800,  double  25,  and  add  to  the 
last  true  divisor  for  the  next  trial  divisor.  This 
method  of  finding  trial  divisors  is-  of  universal 
application,  and  the  rule  may  be  stated  thus: 

Add  to  each  true  divisor,  as  they  occur,  twice  the 
surface  of  one  side  of  the  small  tube,  and  one  each 
of  the  three  parallelopipedons  for  the  trial  divisor  9 
for  that  will  make  three  sides  of  flie  complete  cube. 

We  may  have  an  infinite  number  of  ways  of 
finding  the  cube  root  of  any  number. 

Since  the  cube  root  of  a  number  raised  to  the 
third  power  is  alwafs  equal  to  the  number,  we 
may  presume  the  cube  root  of  1953125  to  be 
divided  into  five  equal  parts  represented  by  5  a. 
The  cube  of  (5a)3  or  125a3  =1953125,  and  a3  wil 
equal  1953125-^125=15625.  The  cube  root  of 
15625  is  25,  a  =25;  5  a=125  the  cube  root  of  the 
number. 


50  HENDERSON'? 

In  the  same  way  we  might  presume  the  root 
to  be  divided  into  25  equal  parts  represented 
by  25a.  25a3  or  15625a3 =1953125,  and  a3  ^125 
and  a =5  and  25a  the  root  of  the  number  equals 
5  times  25  or  125. 

EULE. — Divide  any  number  by  the  cube  of  2,  ex- 
tract tJie  cube  root  of  the  quotient  and  we  have  half 
tliecube  root  of  the  number.  Divide  any  num- 
ber by  the  cube  of  3  or  27,  extract  the  cube  root  of 
the  quotient  and  we  Jiave  one  third  of  the  root  of  the 
number.  Divide  any  number  by  the  cube  of  4 
or  64,  extract  the  cube  root  of  the  quotient  and  we 
have  one  fourth  of  the  root  of  the  number.  Di- 
vide any  number  by  the  cube  of  5  or  125,  extract  the 
cube  root  of  the  quotient  and  we  have  one  fifth  of 
the  root  of  the  number,  etc. 

EXAMPLES  IN  CUBE  BOOT, 

1728)110592(64 

10368 

6912 
6912 

Presume  the  root  to  be  divided  into  twelve 
equal  parts.  Hence  the  cube  root  of  the  quotient 
of  the  number  divided  by  the  cube  of  twelve, 
is  TV  of  the  root  of  the  number. 

The  cube  of  12  is  1728, 110592  -f- 1728  =64,  and 
the  cube  root  of  64  is  4,  4  is  TV  of  the  cube  root, 
the  cube  root  would  be  12  times  4  or  48. 


LIGHTNING  CALCULATOR.  61 

What  is  tlie  cube  root  of  5931 J  ? 

Presume  the  root  to  be  divided  into  13  equal 
parts. 

What  is  the  cube  root  of  117649  ? 

Presume  the  root  to  be  divided  in  7  equal 
parts. 

What  is  the  cube  root  of  97336  ? 

Presume  the  root  to  be  divided  in  23  equal 
parts. 

What  is  the  cube  root  of  95112  ? 

Let  the  root  be  divided  into  29  equal  parts. 

The  number  divided  by  the  cube  of  29  equals 
8,  and  the  cube  root  of  8  is  2.  Hence  29  times 
2  is  the  cube  root  of  the  number  or  58. 

What  is  the  cube  root  of  91125  ? 

Let  the  root  be  divided  into  9  equal  parts,  the 
number  divided  by  the  cube  of  9  equals  125,  the 
cube  root  of  125  is  5.  Hence  9  times  5  or  45  is 
the  cube  root  of  the  number. 

What  is  the  cube  root  of  216x343? 

The  cube  root  of  216  is  6.  The  root  of  343 
is  7. 

The  cube  root  of  the  product  is  6  times  7 
or  42. 

What  is  the  cube  root  of  64x125?; 

The  cube  root  of  64  is  4.  The  cube  root  of 
125  is  5,  5x4=20.  The  cube  root  of  the  pro- 
duct. 

What  is  the  cube  root  of  125x125=5x5  ? 

What  is  the  cube  root  of  125x15625=5x25 


52  HENDERSON'S 

What  is  the  cube  root  of  512x729  ? 

The  cube  root  of  512  is  8.  The  cube  root  of 
729  is  9. 

The  cube  root  of  the  product  is  8x9  or  72. 

What  is  the  cube  root  of  216x729  ? 

The  cube  root  of  216  is  6.  The  cube  root  of 
729  is  9. 

The  cube  root  of  the  product  is  6x9  or  54. 

The  methods  which  we  have  presented  are  of 
universal  application,  and  are  fully  and  clearly 
illustrated  by  Henderson's  book  of  blocks  illus- 
trating roots,  copyrighted  the  llth  clay  of  May, 

A.  D.  1872. 


SQUARE    AND    CUBE  ROOT   OF 
FRACTIONS. 

; 

To  square  a  fraction,  we  square  its  numerator 
for  the  numerator^  and  its  denominator  for  the 
denominator.  Hence,-  to  find  the  square  root 
of  a  fraction,  we  must  extract  the  square  root  of 
its  numerator,  for  the  numerator  of  the  answer, 
and  the  square  root  of  its  denominator  for  the 
denominator  otihe  answer. 

ILLUSTEATIQNS.— Find  the  square  root  of  |. 
The  squ'are  root  of  4,  the  numerator  is  2.  The 
square  root  of  9,  the  denominator,  is  3 .  Hence 
the  answer,  f . 

What  is  the  square  root  of  T2^-  =TV- 


LIGHTNING  CALCULATOR.  53 

What  is  the  square  root  of  .0081  =  .09. 

When  both  terms  of  the  fraction  are  not  per- 
fect squares,  only  an  approximate  Value  of  the 
root  can  be  obtained. 

In  order  that  the  denominator  of  a  decimal 
fraction  may  be  a  perfect  square,  its  numerator 
must  contain  an  even  number  of  decimal  places. 
Hence,  to  extract  the  square  root  of  a  decimal 
fraction,  make  its  number  of  decimal  placqs 
even,  by  annexing  a  zero,  if  necessary;  extract 
the  root,  as  in  whole  numbers,  observing  that 
there  will  be  one  decimal  place  in  the  root  for 
'every  two  hi  the  given  fraction,  the  root  may  be 
found  to  any  number  of  decimal  places  by  an- 
nexing two  zeros  for  every  additional  figure. 

To  extract  the  cube  root  of  a  fraction,  we  ex- 
tract the  cube  root  of  the  numerator  for  the 
numerator  of  tlie  answer,  and  the  cube  root , of 
the  denominator  for  the  denominator  of  the  an- 
swer. If  its  numerator  and  denominator  are 
not  perfect  cubes,  the  approximate  value  of  the 
cube  root  can  only  be  obtained.  If  the  denom- 
inator is  not  a  perfect  cube,  both  terms  should 
be  multiplied  by  the  square  of  the  Denominator. 
Hence,  to  extract  the  square  root  of  a  decimal 
fraction,  annex  zeros,  if  necessary,  to  make. its 
number  of  decimal  places  some  multiple  of 
three;  extract  its  root,  as  in  whole  numbers, 
observing  that  there  will  be  one  decimal  place 
for  everythree  in  the  given  fraction. 


54  HENDERSON'S 

TO    FIND    THE    SURFACE    OF 
PLANE    FIQUKES. 

A  triangle  is  a  figure  having  three  sides  and 
three  angles. 

The  altitude  of  a  triangle  is  the  perpendicular 
distance  from  the  side  assumed  as  its  base  to 
to  the  vertex  of  the  opposite  angle.  B  c  is  the 
perpendicular,  and  the  A  D  the  base. 

KULE. — To  find  the  surface  of  any  triangle, 
multiply  the  base  by  half  the  altitude. 

A  right-angle  triangle  is  a  triangle  having  a 
right  angle. 

Lines  are  parallel  when  they  lie  in  the  same 
direction.  A  parallelogram  is  a  fotir-sided  figure 
having  its  opposite  sides  parallel. 

A  trepizoid  is  a  four-sided  figure,  having  two 
of  its  sides  parallel. 

A  polygon  is  a  figure  bounded  on  all  sides 
by  straight  lines. 

Similar  figures  are  those  which  have  the  same 
shape. 

The  corresponding  sides  are  proportional. 

The  base  of  a  figure  is  the  side  on  which  it  is 
supposed  to  stand. 

The  altitude  of  a  rectangle,  a  parallelogram 
or  a  trepizoid,  is  the  perpendicular  distance 
between  its  parallel  basis. 

The  area  of  a  rectangle  is  4he  length  multi- 
plied by  the  width. 


UGHTNING  CALCULATOR.  55 

METHOD  OF  MEASUBING  LAND. 

Find  the  number  of  rods  by  multiplying  the 
length  by  the  width.  Kemove  the  point  two 
places  to  the  left,  divide  by  eight  and  multiply 
the  quotient  by  five;  or  remove  the  point  two 
places,  take  f  of  the  result,  and  we  have  the 
number  of  acres.  Thus:  3280  rods,  the  point 
removed  two  places  leaves  32.80-5-8  =  4.1. 
4,1  X  5  =  20.5  acres. 

What  is  the  number  of  acres  in  2440  rods  ? 
Remove  the  point  two  places  we  have  24.40;  f 
X  24.40  is  15J,  the  number  of  acres.  This 
method  is  of  universal  application,  and  may  be 
stated  in  the  following  words :  Remove  the  deci- 
mal point  two  places  to  tlie  left,  and  f  of  the  quotient 
are  tlie  number  of  acres. 

We  remove  the  point  two  places  to  reduce  the 
number  to  units  of  a  hundred,  and  since  there 
are  f  of  a  hundred  rods  in  one  acre,  five  times 
J  of  the  number  of  hundred  rods  must  equal  the 
number  of  acres;  or  simply  ftie  point  removed 
two  places  and  the  quotient  divided  by  f  equals 
the  number  of  acres. 

What  are  the  number  of  acres  in  a  field  160 
rods  wide  and  480  rods  long?  Kemove  the 
point  two  places  on  160,  and  take  f  of  the  quo- 
tient, we  find  one  acre  multiplied  by  480,  the 
length,  we  get  480  acres,  Ans. 

What  is  the  number  of  acres  in  a  field  2200 
rods  long  and  640  wide? 


56  HENDERSON'S 

What  is  the  number  of  acres  in  a  field  of  tri- 
angular shape  ?  The  base  of  the  triangle  is  800 
rods  and  the  altitude  300;  since  the  area  is  the 
base  multiplied  by  half  the  altitude.  Half  the 
altitude  is  150;  remove  the  point  two  places  on 
800,  and  we  have  8,  and  f  X8=5,  and  5x150= 
750,  the  number  of  acres  in  the  field. 

The  area  of  a  circle  also  equals  the  square  of 
its  radius  multiplied  by  3.1416,  the  ratio  of  the 
circumference  to  the  diameter.  If  the  radius  is 
two  feet  the  area  of  the  circle  is  3.1416x22= 
12.5664. 

Find  the  area  of  a  circle  12  feet  in  diameter. 

Find  the  area  of  a  circle  of  8  feet  radius;  of 
a  circle  of  100  feet  radius. 

The  suface  of  a  sphere  equals  the  square  of 
its  diameter  multiplied  by  3.1416. 

ILLUSTRATION. — The  surface  of  a  sphere  5  feet 
in  diainetei— 3. 1416x25. 

The  surfaces  of  spheres  are  to  each  other  as 
the  squares  of  their  diameters. 

The  solidity  of  a  sphere  equals  the  product 
of  the  surface  multiplied  by  ^  of  the  diameter, 
or  it  equals  ^  of  the  cube  of  the  diameter  mul- 
tiplied by  3.1416.  The  solidities  of  spheres  are 
to  each  other  as  the  cubes  of  their  diameters^ 

The  solidities  of  similar  solids  arc  to  each, 
other  as  the  cubes  of  their  like  dimensions. 

The  solidity  of  a  cylinder  equals  the  product 
of  the  area  of  its  base  by  its  altitude. 


OF  THB 

UNIVERSITY 


LIGHTNING  CALCULATOR. 

The  convex  surface  of  a  cylinder  equals  tlic 
product  of  the  circumference  of  its  base  by  its 
altitude. 

What  is  the  solidity  of  a  cylinder  8  feet  higb 
with  a  base  4  feet  in  diameter?  A  cylinder  15 
feet  high,  with  a  base  1  foot  diameter? 

What  is  the  diameter  of  a  sphere  containing 
100  cubic  feet? 

One  bushel  is  about  f  of  a  cubic  foot.  Hence 
-f  of  the  number  of  cubic  feet  equals  the  num- 
ber of  bushels  nearly.  The  dimensions  of  a 
box  are  12  feet  long,  6  feet  in  width,  and^  5  feet 
high.  How  many  bush  ols  does  it  contain?  The 
product  of  12,  6  and  5  is  360;  Number  of  cubic 
feet=288  bushels.  Eemove  the  point  one  place 
to  the  left  and  multiply  by  8. 

Hence  the  rule  to  find  the  number  of  bushels 
from  the  number  of  cubic  feet.  ,. 

Remove  the  decimal  point  one  place  to  the  left, 
and  multiply  the  quotient  by  8. 

ILLUSTRATION;—  In  a  bin  of  800.9  cubic  feet, 
remove  the  point  one  place  to  the  left,  we  have 
80.09;  multiply  by  8  and  we  have.  640.  72;  the 
number  of  bushels. 

To  find  the  number  of  cubic  feet  from  the 
bushels,  simply  increase  tho  number  by.  olio 
quarter  of  itself. 

What  is  the  number  of  bushels  that  a  bin  will 
contain,.  20  feet  long,  8  wide,  5J  deep? 
What  is  the  number  of  cubic  feet  in  2150  bushels  2 


68  HENDERSON'S 

SOME  OF  THE  MISCELLANEOUS  WEIGHTS  TO  THE 

BUSHEL. 


60  !bs  make  1  bushel  of  Wheat. 

56        "         1        "           Corn. 

33        "         1        "           Oats. 

£8        "         1 

Barley.  . 

66        "         1 

'           Bye. 

60        "         1 

*           Beans. 

52        "         1 

'          Buckwheat. 

70 

1 

*           Corn  in  ear. 

50 

1 

'           Corn  meal. 

60 

i 

*          Potatoes. 

50 

1        "           Salt. 

33 

'         1        "          Peaches,  dried. 

25 

1        "          Apples,  dried. 

62 

'         1        "           Clover  seed. 

45 

*         1        "           Timothy. 

56 

1        "          Flax. 

SHORT    METHODS    IN    DIVIS- 
ION AND  MULTIPLICATION. 

Remove  the  point  one  place  to  the  right  to 
multiply  by  10;  two  places  to  multiply  by  100; 
three  places  1000,  etc. 

To  divide,  remove  it  to  the  left. 

To  multiply  by  25,  divide  by  4  and  call  the 
quotient  hundreds. 

Thus:  25x480—12000.  480-^4=120  call  it 
hundreds,  makes  12000.  Divide  by  4,  be- 
cause 25  is  one  quarter  of  a  hundred. 

To  multiply  by  2|  divide  by  4  and  call  it  tens; 
call  it  tens,  because  2J  is  the  quarter  of  ten. 


LIGHTNING  CALCULATOR.  69 

To  multiply  by  125,  divide  by  8  and  call  it 
thousands.  Call  it  thousands,  because  125  is  J 
of  a  thousand. 

To  multiply  by  12  J  divide  by  8;  call  it  hun- 
dreds. 

To  multiply  by  1£  divide  by  8;  call  it  tens. 

To  multiply  by  62|  divide  by  16  and  call  it 
thousands. 

To  multiply  by  6J  divide  by  16  and  call  it 
hundreds. 

To  multiply  by  31£  divide  by  32  and  call  it 
thousands. 

To  multiply  by  333£,  divide  by  3  and  call  it 
thousands. 

To  multiply  by  33  J,  divide  by  3  and  call  it 
hundreds. 

To  multiply  by  3  J,  divide  by  3  and  call  it  tens. 

To  multiply  by  50,  divide  by  2  and  call  it  hun- 
dreds. 

To  multiply  by  66f ,  divide  by  15  and  call  it 
thousands. 

To  multiply  oy  6f ,  divide  by  15  and  call  it 
hundreds. 

To  multiply  by  833J,  divide  by  12  and  call  it 
ten  thousands,  by  annexing  four  ciphers. 

To  multiply  by  83 J,  divide  by  12  and  call  it 
thousands. 

To  multiply  by  8J,  divide  by  12  and  call  it 
hundreds.  Divide  by  12  and  -call  it  hundreds. 


60  EENDEKSON'S 

because  8J-  is  -^  of  a  hundred.  The  reason  is 
similar  in.  .each  case. 

The  primitive  meaning  of  reason  is  hook  some- 
thing to  hold  on  by.  Please  get  the  reason  in 
each  case. 

To  multiply  by  166f,  divide  by  6  and  call  it 
thousands;  because  166|  is  -f  of  1000. 

To  multiply  by  16| ,  divide  by  6  and  call  it 
hundreds. 

To  multiply  by  If,  divide  by  6  and  call  it  tens. 

To  multiply  by  37 J,  take  f  of  the  number  and 
call  it  hundreds;  87  J,  ^  of  the  number,  and  call 
it  hundreds,  etc. 

We  simply  reverse  these  methods  to  divide. j 

To  divide  by  10,  100,  1000,  etc.,  we  remove 
tte  point  one,  two,  and  three  places  to  the  left. 

To  divide  by  25,  remove  the  decimal  point 
two  places  to-the  left  and  multiply  by  4. 

Eemoving  the  point  two  places  divides  by  one 
hundred;  hence  the  (|uotient  is  4  times  to  small; 
kence  we  remove  the  point  two  places  $nd 
multiply  by  4,  :  •  r 

To  divide  by  2J,  remove  the  point  one  place 
to  the  left  and ^multiply  by  4. 

To  divid^  ? by  125,  remove  the  point  three 
places  to  the  left  a:hd  multiply  by  8. 

To  divide  by  12  J,  remove  the  point  two  places 
to  the  left  and  multiply  fey  8.  -  < 

L  To  divide  by  1| -,  remove  the  point  one  place 
to  the  left  and  multiply  by  8.  There  are  about 


LIGHTNING  CALCULATOB.  61 

1J  cubic  feet  in  one  bushel.  Hence  divide  the 
number  of  cubic  feet  by  1J  gives  the  number  of 
bushels  nearly. 

To  divide  by  625,  remove  the  point  four  places 
to  the  leffc  and  multiply  by  16. 
.   To   divide   by  62J,   remove  the   point  three 
places  to  the  left  and  multiply  by  16. 

To  divide  by  6 J,  remove  the  point  two  places 
to  the  left  and  multiply  by  16. 

To  divide  by  3125,  remove  the  point  five  places 
to  the  left  and  multiply  by  32. 

To  divide  by  3J,  remove  the  point  two  places 
to  the  left  and  multiply  by  32. 

To  divide  333  J,  remove  the  point  three  places 
to  the  left  and  multiply  by  three. 

To  divide  by  666|,  remove  the  point  four 
places  to  the  left  and  multiply  by  15. 

To  divide  by  66|,  remove  the  point  three 
places  to  the  left  and  multiply  by  15. 

To  divide  by  833 J,  remove,  khe  point  four 
places  to  the  left  and  multiply  by  12.  - 

To  divide  by  83^,  remove  .fhe  point  three 
places  to  the  left  and  multiply  by  12. 

To  divide  by  8^,  remove  the  point  two.  places 
to  the  left  and  multiply  by  12. 

To  divide  by  166f ,  remove  the  point  three 
places  to  the  left  and  multiply  by  6.  Eemoving 
the  point  three  places  divides  by  1000;  hence 
the  quotient  is  6  times  too  small.  166|  is  -J-  of 
1000. 


62  HENDERSON'S 

MENTAL  EXERCISE. 

PROBLEM  1. — Take  1,  multiply  by  49,  extract 
the  square  root,  multiply  by  4,  subtract  1,  and 
extract  the  cube  root;  what  is  the  result? 

PROBLEM  2. — Take  9,  divide  by  2,  multiply  by 
6,  extract  the  cube  root,  multiply  by  27,  and 
extract  the  fourth  root;  what  is  the  result? 

PROBLEM  3.— T&ke  48,  divide  by  2,  multiply 
by  4,  add  4,  extract  the  square  root,  multiply 
by  5,  subtract  1,  divide  by  seven,  and  what  is 
the  result? 

PROB'.^^  4.— Take  8|,  multiply  by  8J,  sub- 
tract f,  divide  by  8,  extract  the  square  root, 
multiply  by  40  a&d  divide  by  10;  what  is  the 
result? 

PROBLEM  5. — Take  1J,  multiply  by  1J,  2|  by 
2J,  3|  by  3f ,  run  it  up  to  12J,  in  concert. 

PROBLEM  6.— Take  1£,  multiply  by  If ,  2  J  by 
2|,  etc.,  up  to  12. 

PROBLEM  7.— Take  If,  multiply  by  If,  2f  by 
2|,  etc.,  up  to  15. 

PROBLEM  8.— Take  If,  multiply  by  l£,  2f  by 
2£,  etc.,  up  to  20. 

PROBLEM  9.— Take  If,  multiply  by  If,  2f  by 
2f,  etc.,  up  to  17. 

PROBLEM  10.— Take  If,  multiply  by  If,  2f  by 
2|,  etc. 

PROBLEM  11.— Take  1  \,  multiply  by  1TV,  2T5? 
by  2&,  etc. 


LIGHTNING  CALCULATOB.  68 


PROBLEM  12.—  Take  1^,  multiply  by  lT4f,  2TV 
by  2tV  etc. 

PROBLEM  13.—  Take  12J,  multiply  by  12J,  11* 
by  11  J,  etc.,  down  to  1. 

PROBLEM  14.—  Take  11J,  multiply  by  llf,  10J 
by  10|,  etc.,  down  to  1. 

PROBLEM  15.—  Take  12£,  multiply  by  12J,  11£ 
by  11£,  etc.,  down  to  1. 

PROBLEM  16.—  Take  13f  ,  multiply  by  13f  ,  12| 
by  12|,  etc.,  down  to  1. 

PROBLEM  17.—  Take  12TV,  multiply  by  12TV, 
11TV  by  H^V  e*c-»  down  to  1. 

PROBLEM  18.—  Take  10T\,  multiply  by  10^, 
etc.,  down  to  1. 

PROBLEM  19.—  Take  12T\,  multiply  by  12&, 
etc.,  down  to  1. 

PROBLEM  20.—  Take  8^,  multiply  by  8T\, 
7^  by  7T\,  etc.,  down  to  1. 

PROBLEM  21.—  Take  10TV,  multiply  by 

9ir  ^J  ^TT*  e*c->  down  to  1. 

PROBLEM  22.—  Take  12-&,  multiply  by 
etc.,  down  to  1. 

PROBLEM  23.—  Take  !!&,  multiply 
etc.,  down  to  1. 

PROBLEM  24.—  Take  12|,  multiply  by  12|,  etc., 
down  to  1. 

PROBLEM  25.—  Take  8T\,  multiply  by  8j|,  etc., 
down  to  1. 

PROBLEM  26.—  Take  13|,  multiply  by  13^;  etc., 
down  to  1. 


64  HENDERSON'S 

V 

The  mean  of  two  numbers,  is  half  their  sum,  or 
the  number  equally  distant  from  the  two  numbers. 

The  product  of  two  numbers  is  the  square  of  their 
mean  diminished  by  the  square  of  half  of  their  dif- 
ference. 

PROBLEM  27. — 19  times  21,  18  times  22,  etc., 
down  to  15.  Tims  :  The  mean  is  20,  the  square 
of  20  is  400,  400— the  square  of  1  is  399 ;  the 
product,  18  times  22  is  the  square  of  20,  400 — 
the  square  of  2,  4,  396.  17  times  23  is  391,  16 
times  24,  384 ;  15  times  25,  375. 

PROBLEM  28.— Take  29  by  31,  28  by  32,  etc., 
down  to  20  and  up  to  40. 

PROBLEM  29.— Take  39  by  41,  38  by  42,  etc., 
down  to  30  and  up  to  50. 

PROBLEM  30.— Take  49  by  51,  48  by  52,  etc., 
down  to  40  and  up  to  60. 

PROBLEM  31.— Take  59  by  61,  58  by  62,  etc., 
dowm  to  50  and  up  to  70. 

PROBLEM  32.— Take  69  by  71,  68  by  72,  etc., 
down  to  60  and  up  to  80. 

PROBLEM  33.— Take  79  by  81,  78  by  82,  etc., 
down  to  70  and  up  to  90. 

*     PROBLEM  34.— Take  89  by  91,  88  by  92,  etc., 
down  to  90  and  up  to  100. 

The  complement  of  a  number  is  the  difference  of 
that  number  and  some  particular  number  above 
it.  The  supplement  of  a  number  is  the  difference 
of  that  number  and  tome  particular  number  be- 


LIGHTNING-  CALCULATOR  65 

Thus,  the  complement  of  99  is  the  difference 
of  99  and  100,  which  is  1. 

The  supplement  of  101  is  the  difference  of 
101  and  100,  which  is  1. 

PEOBLEM  35.  — Commence  at  99  and  square  num- 
bers down  to  90.  Thus  :  99  times  99  is  9801, 


98  times  98  is  9604,  97  times  97  is  9409,  96 
times  96  is  9216,  etc.  Simply  diminish  the 
number  by  its  complement,  call  it  hundreds  and 
add  the  square  of  the  complement. 

When  we  use  the  supplement,  wre  add  it  to 
the  number,  give  it  its  proper  name  and  add 
the  square  of  the  supplement. 

Thus  :  101  times  101,  the  supplement  1  added 
to  101  makes  102,  call  it  hundreds,  is  10200, 
plus  the  square  of  tho  supplement  is  10201. 

PROBLEM  36. — Commence  at  101,  square  all 
the  numbers  up  to  110  and  down  to  90. 

PEOBLEM  37. — Commence  at  51,  square  all  the 
numbers  up  to  60  and  clown  to  40. 

PROBLEM  38. — Commence  at  21,  square  all 
the  numbers  up  to  25  and  down  to  15, 

PEOBLEM  39. — Commence  at  11,  square  all 
the  numbers  up  to  15  and  down  to  5. 

PEOBLEM  40. — Commence  at  999,  square  all 
the  numbers  down  to  990  and  up  to  1010,  etc.. 
etc.,  etc.,  etc. 


66  HENDERSON'S 


MISCELLANEOUS    PROBLEMS. 

PROBLEM  1. — How  many  bushels  in  a  bin  10 
feet  long,  4  feet  wide  and  4  feet  deep? 

SOLUTION. — Since  there  are  ^°  of  a  cubic  foot 
in  one  bushel,  the  bin  will  contain  8  times  TV 
of  the  number  of  cubic  feet,  in  bushels.  TV  of 
10  is  1,  8  times  1  are  8,  4  times  8,  32,  and  4" 
times  32,  128,  Ans.  Or  find  the  number  of 
cubic  feet  in  the  bin,  remove  the  decimal  point 
one  place  to  the  left,  and  multiply  by  8  in  all 
cases.  Thus  :  the  product  of  4,  4  and  10  is  160; 
remove  the  point  one  place  to  the  left  and  we 
have  16,  16  multiplied  by  8  is  128,  Ans. 

PROBLEM  2. --How  many  bushels  in  a  bin  32 
feet  long,  16  feet  wide  and  5J  feet  high? 

PROBLEM  3. — How  many  bushels  in  a  bin  24 
f*et  long,  12  feet  wide,  4J  feet  high? 

PROBLEM  4. — A  cubic  foot  of  water  weighs  62 
Ibs.  8  oz. — what  is  the  pressure  011  5  acres  at 
the  bottom  of  the  sea,  where  the  water  is  1  mile 
deep  ? 

PROBLEM  5. — What  would  be  the  weight  of 
this  planet  if  one  cubic  foot  weighs  62-|  pounds? 

PROBLEM  6. — If  21|  bushels  of  oats  are  re- 
quired to  seed  9f  acres,  how  many  bushels  will 
be  required  to  seed  a  field  of  100  acres? 

PROBLEM  7.— If  33 £  pounds  of  tea  cost  $27  J, 
Low  much  will  300  pounds  cost? 


LIGHTNING  CALCULATOR.  67 

PROBLEM  8. — A  field  3J  times  as  long  as  it  is 
wide  contains  30  acres — what  are  its  dimensions? 

PROBLEM  9. — If  each  one  of  20  pupils  breathe 
30  cubic  feet  of  air  per  hour,  in  how  long  a 
time  will  they  breathe  as  much  air  as  a  room  20 
by  30  and  8  feet  high  contains? 

PROBLEM  10. — If  gold  is  1.12J,  what  is  cur- 
rency worth? 

SOLUTION. — The  value  of  currency  would  be 
TI¥K»  simply  multiplying  the  numerator  and 
denominator  lyy  2  and  we  have  ||f  =  f  ;  hence 
one  dollar  in  currency  is  worth  f  X  1y°  cents,  or 
88 1  cents. 

PROBLEM  11. — If  currency  is  worth  88f  cents 
on  the  dollar,  what  is  gold  worth?  Simply  in- 
vert the  preceding  operation. 

PROBLEM  12. — If  gold  is  1.10J,  what  is  cur- 
rency ? 

PROBLEM  13. — If  currency  is  95  cents  on  the 
dollar,  what  is  gold? 

PROBLEM  14. — If  a  wolf  can  eat  a  sheep  in  -J 
of  an  hour,  and  a  bear  in  f  of  an  hour,  how 
long  will  it  take  them  together  to  eat  what  re- 
mains of  a  sheep  after  the  wolf  has  been  eating 
half  an  hour? 

SOLUTION. — In  one  hour  the  wolf  eats  f  of  a 
sheep,  after  eating  half  an  hour  f  of  the  sheep 
would  remain,  since  in  one  hour  they  eat  -§— (- f  or 
|f;  to  eat  -|  or  ¥°r  of  a  sheep  it  would  take  them 
as  long  as  |f  is  contained  in  -^y,  which  is  -£%  of 
an  hour,  Ans. 


68  HENDERSON'S 

PROBLEM  15. — John  cuts  a  cord  of  wood  in  f 
of  a  day,  James  in  f  of  a  day,  how  long  will  it 
take  them  to  cut  a  cord  when  they  work  to- 
gether? 

PROBLEM  16. — A  can  do  a  piece  of  work  in  8 
days  and  A  and  B  can  do  the  same  in  5  days ; 
after  A  did  ^  of  the  work,  B  did  the  remainder 
— how  long  did  it  take  him? 

PROBLEM  17. — Divide  the  number  108  into' 
t\ro  such  parts,  that  f  of  the  first-j-8  shall  equal 
the  second. 

PROBLEM  18. — A  ship  mast  63  feet  in  length, 
in  a  storm,  was  broken  off;  f  of  what  was 
broken  off  equaled  f  of  what  remained;  how 
much  was  broken  off,  and  how  much  remained? 

PROBLEM  19. — A  farmer  has  2290  sheep  in 
two  fields,  f  of  the  number  in  the  first  field 
equals  f  of  the  number  in  the  second;  how 
many  are  there  in  each  field? 

PROBLEM  20. — A  market  woman  was  requested 
to  buy  99  fowls,  consisting  of  two  different 
kinds;  J  of  the  number  of  the  first  kind  was  to 
equal  |  of  the  second  kind;  how  many  of  each 
kind  must  she  buy? 

PROBLEM  21. — A  farmer,  after  selling  f  of  1-J 
times  as  much  grain  as  he  had,  had  100  bushels 
remaining:  how  much  had  he  at  first? 

PROBLEM  22. — Divide  the  number  170  into 
two  parts,  that  shall  be  to  each  other  as  §  to  f . 

PROBLEM  23. — |  of  A's  number  of  sheep  plus 


LIGHTNING  CALCULATOB.  09 

f  of  B's  number  equals  900;  how  many  sheep 
has  each,  providing  f  of  B's  number  is  4-  of  A's 
number: 

PEOBLEM  24. — A  gold  and  silver  watch  were 
bought  for  $320;  the  silver  watch  cost  T  as 
much  as  the  gold  one;  what  was  the  cost  of 
each  ? 

PROBLEM  25 — J  of  A's  money  -f~  §  of  B's; 
equals  6800;  and  f  of  B's  is  4  times  -J-  of  A's; 
how  much  money  has  each? 

PROBLEM  26. — Divide  the  number  60  into  two 
parts,  that  shall  be  to  each  other  as  -J  to  f 

PEOBLEM  27. — The  sum  of  two  numbers  is 
140,  and  the  larger  is  to  the  smaller  as  1  to  |-; 
what  are  the  numbers  ?\ 

PEOBLEM  28. — A  and  B  together  owe  $207; 
B  owes  -i^-  as  much  as  A;  how  much  does  each 
owe  ? 

PEOBLEM  29. — I  sold  a  horse  for  J  more  than 
he  cost  me,  receiving  $270  for  him;  how  much 
did  he  cost  me? 

PEOBLEM  30. — What  will  f  of  a  barrel  of  flour 
cost  at  $11.28  per  barrel? 

PEOBLEM  31. — What  will  |  of  a  bag  of  coffee 
weigh  if  a  bag  weighs  147  BbB? 

PEOBLEM  32. — What  will  |  of  a  pound  of  tea 
cost  at  $1.25  per  pound? 

PEOBLEM  33. — What  will  T°¥  of  a  cord  of  wood 
cost  at  $6.25  per  cord? 

PEOBLEM  34. — What  will  TV  of  a  hogshead  of 
wine  cost  at  $138.75  per  hogshead? 


70  HENDERSON'** 

PROBLEM  35. — How  much  is  |  and  J  of  -J-  of  15? 

PROBLEM  36. — A  and  B  traded  in  company; 
A  piji  in  §  as  much  as  B;  they  gained  $750; 
what  was  each  man's  share? 

PROBLEM  37. — James  says  to  John,  give  me 
87.00  and  I  will  have  as  much  money  as  you. 
John  says  to  James,  give  me  $7.00  and  I  will 
have  twice  as  much  as  you,  Ans.  35  and  49. 

Simply  multiply  the  $7.00  by  the  numbers  5 
and  7 ;  and  for  all  similar  problems  simply  mul- 
tiply the  sum  of  money  given,  by  the  numbers 
5  and  7. 

PROBLEM  38. — A  says  t&  B,  give  me  $3J  and  I 
will  have  as  much  money  as  you.  B  says  to  A, 
give  me  $3J-  and  1  will  have  twice  as  much  as 
you.  How  much  money  has  each? 

PROBLEM  39. — Haight  says  to  Booth,  give  me 
1000  sheep  and  I  will  have  as  many  as  you. 
Booth  says  to  Haight,  give  me  1000  and  I  will 
have  twice  as  many  as  you.  How  much  has  each? 

PROBLEM  40. — Friedlander  says  to  Eeese,  give 
me  $500,000  and  I  will  have  as  much  as  you. 
Eeese  says  to  Friedlander,  give  me  $500,000 
and  I  will  have  twice  as  much  as  you.  How 
much  has  each? 

PROBLEM  41. — G  says  to  D,  give  me  $13.33J 
and  I  will  have  as  much  money  as  you.  D  says 
to  G,  give  me  13.33J  and  I  will  have  twice  as 
much  money  as  you.  How  much  has  each? 

PROBLEM  42. — Greeley  says  to  Grant,  give  mo 


LIGHTNING  CALCtJLATOB.  71 

50,000  votes  and  I  will  Lave  as  many  as  yon. 
Grant  says  to  Greeley,  give  me  50,000  votes 
and  I  will  have  twice  as  many  as  you;  how  many 
has  each?  A 

PROBLEM  43. — Two  Hoodlums  go  imko  a  saloon; 
one  says  to  the  other,  give  me  as  much  money 
as  I  have,  and  I  will  spend  two  bits  with  you. 
They  go  into  another  saloon,  and  he  sajs,  give 
me  as  much  money  as  I  now  have,  and  I  vail 
spend  two  bits  with  you.  They  went  into  the 
third  saloon,  and  he  made  the  same  statement, 
and  when  they  came  out  of  the  third  saloon  he 
had  nothing  left.  How  much  had  he  when  he 
went  into  the  first  saloon?  Ans.,  If  bits. 
Simply  -£  of  the  sum  borrowed  in  the  first  saloon 
is  the  answer. 

PROBLEM  44. — A  and  B  step  into  a  hotel;  A 
says  to  B,  give  me  as  much  money  as  I  have, 
and  I  will  spend  five  dollars  with  you.  They 
go  into  a  second  and  third,  A  making  the  same 
statement;  and  when  they  came  out  of  the  third, 
he  had  nothing  left.  How  much  had  he  when 
they  went  into  the  first  hotel  ? 

PROBLEM  45.— If  3  be  the  third  of  6.  what 
will  the  fourth  of  20  be?  Ans.  3J. 

SOLUTION.— The  third  of  6  is  2,  if  3  be  2,  1 
is  J  of  2  or  f,  and  20  is  20  times  §  or  4^,  the  -J 
of  20  is  the  J  of  V  or  V°,  3J-  Ans. 

PROBLEM  46.— If  the  third  of  6  be  3  what  will 
the  fourth  of  20  be?  Ans.  7i. 


72  HENDERSON'S 

SOLUTION.— If  2  be  3,  1  is  £  of  3,  1J  and  2Q 
is  20  times  1J  or  30,  i  of  20  is  the  £  of  30  or 
7*  Ans. 


••* 
GMENTERAL,    IXFORY1ATION. 

The  circumference  of  a  circle  equals  the  diam- 
eter multiplied  by  3.1416,  the  ratio  of  the  cir- 
cumference to  tho  diameter. 

The  radius  of  a  circle  equals  the  circumfer- 
ence multiplied  by  6.283185. 

The  area  of  a  circle  equals  the  square  of  the 
radius  multiplied  by  3.1416. 

The  area  of  a  circle  equals  tiie  square  of  the 
diameter  multiplied  by  7854. 

The  area  of  a  circle  equals  one  quarter  of  the 
diameter  multiplied  by  the  circumference. 

The  radius  of  a  circle  equals  the  circumfer- 
ence multiplied  by  0.159155. 

The  radius  of  a  circle  equals  the  square  root 
of  the  area  multiplied  by  0.56419. 

The  diameter  of  a  circle  equals  the  •  circum- 
ference multiplied  by  0.31831. 

The  diameter  of  a  circle  equals  the  square 
root  of  the  area  multiplied  by  1.12838. 

The  side  of  an  inscribed  equilateral  triangle 
equals  the  diameter  of  the  circle  multiplied  by 
0.86. 

The  side  of  an  inscribed  square  equals  the 
diameter  multiplied  by  0.7071. 


LIGHTNING  CALCULATOR.  73 

The  side  of  an  inscribed  square  equals  the 
diameter  of  the  circle  multiplied  by  0.225. 

The  circumference  of  a  circle  multiplied  by 
0.282  equals  one  side  of  a  square  of  the  same 
area. 

The  side  of  a  square  equals  the  diameter  of  a 
circle  of  the  same  area  multiplied  by  0.8862. 

The  area  of  a  triangle  equals  the  base  multi- 
tiplied  by  one  half  of  its  altitude. 

The  area  of  an  ellipse  equals  the  product  of 
both  diameters  and  .7854. 

The  solidity  of  a  sphere  equals  its  surface 
multiplied  by  one-sixth  of  its  diameter. 

The  surface  equals  the  product  of  the  diam- 
eter and  circumference. 

The  surface  of  a  sphere  equals  the  square  of 
the  diameter  multiplied  by  3.1416. 

The  surface  equals  the  square  of  the  circum- 
ference multiplied  by  0.3183. 

The  solidity  of  a  sphere  equals  the  cube  of 
the  diameter  multiplied  by  0.5236. 

The  diameter  of  a  sphere  equals  the  square 
root  of  the  surface  multiplied  by  0.56419. 

The  square  root  of  the  surface  of  a  sphere 
multiplied  by  1.772454  equals  the  circumference. 

The  diameter  of  a  sphere  equals  the  cube  root 
of  its  solidity  multiplied  by  1.2407. 

The  circumference  of  a  sphere  equals  the 
cube  root  of  its  solidity  multiplied  by  3.8978. 

The  side  of  an  inscribed  cube  equals  the 
radius  multiplied  by  1.1547. 


74  HENDEESON'S 

The  solidity  of  a  cone  or  pyramid  equals  the 
area  of  its  base  multiplied  by  one  third  of  its 
altitude. 


75  LIGHTNING   CALCULATOR. 

COLLEGE  DE  L'UNION. 

DIPLOME  DE  BACHELIER  DBS  ARTS. 
Nous  Directeurs  du  College  de  TUnion  a  Sche- 
nectady,  Etat  de  New  York,  vu  le  Certificat  cV apti- 
tude au  grade  de  Bachelier  es  Arts,  accorde  par  la 
Faculte  du  College  au  Sieur  Jean  Alexandre  Hen- 
derson, ratinant  le  susdit  Certincat.  Donnons  par 
ces  presentes  au  dit  Sieur,  le  Diplonie  de  Bachelier 
es  Arts, pour  en  jouir  avec  les  droits  et  prerogatives 
qui  y  sont  attaches.  En  teinoignage  de  quoi  nous 
avons  muni  ce  Diplome  de  notre  sceau  et  des  sig- 
natures du  President  et  des  Professeurs  de  ce  Col- 
lege. 

Fait  &  Schenectady  le    vingt  huitieme  Juillet 
1864, 

E.  NOTT,  Pres< 
L.  P.  HICKOK,  Acting  Pres. 
J.  H.  JACKSON,  Prof,  de  Math. 
JOHN  FOSTER,  Prof,  de  Physique. 
GUILL.  M.  GTLLESPIE, 

Prof,  de  Fonts  et  Chaussees. 
C.  F.  CHANDLER,  Ghem.  Prof. 
W.  LAMOREUX, 

Acting  Prof.  Lang.  Mod. 
N.  G.  CLARK, 

Prof,  des  Belles  Lettres. 
JONATHAN  PEARSON, 

Prof.  Hist.  Nat. 

N.  B. — J.  A.  Henderson  a  regu  le  A.  M,  degre 
de  Maitre,  1867. 


LE  CALCULATEUR  INSTANTANE. 


PAR  J.  JL.-  HENDERSON. 


L'alphabet  arithmetique  est~e*crit  et  lu:  Le  se- 
cond  est  deux 

a  „          fois  le  premier, 

a  3  i  9  §  3  j  .g  ^  6  le  troisieme 
&  g  .2  H  2*  ^  •&  ^S  e^a  troisfoislepre- 
§r§|&.S.y&Vg  '  mier,etainside 

1234567890      suite  iusqu'a  la 

&« 
-•    fin. 

1111111111 

INTfiEET. 

Comme  Tinte're^t  est  gen^ralement  une  portion 
du  principal,  la  me'thode  de  le  calculer  viendra 
sur  la  methode  de  le  diviser.  La  regie  e*tablira  le 
temps  quand  la  piastre  fait  un  cent,  et  nous  place- 
rons  le  chiffre  decimal  deux  places  a  la  gauche,  parce- 
que  un  centieme  du  principal  egal  Tinteret.  Dans 
dix  fois  le  temps  la  piastre  fait  dix  cents,  et  nous 
placerons  le- chiffre  decimal  une  place  a  la  gauche, 
parceque  un  dixieme  du  principal  en  est  Tinteret; 


LIGHTNING   CALCULATOR. 


77 


dansundixieme  du  meme  temps  une  piastre  fait  un 
mille,  et  nous  placerons  le  chiffre  decimal  trois 
places  a  la  gauche,  parceque  un  millieme  du  prin- 
cipal en  est  Tinte're^t. 

REGLE — Le  reciproque  du  prix,  ou  le  prix  ren- 
verse indique  le  temps  quand  nous  pourrons  chan- 
ger le  chiffre  decimal  deux  places  a  la  gauche;  en 
tout  cas,  dix  fois  le  temps  une  place  a  la  gauche, 
et  un  dixieme  du  meme  temps  trois  place  a  la 
gauche,  augmente  ou  diminue  le  resultat  afin  de 
retrouver  le  temps. 

L'alphabet  numerical  est  \,  $,  ?,  1,  f ,  ?,  I,  !,  ?, 
etO;  I  renverse  est  1;  ?  renverse'  est  ^;  ^enverse  J, 
etc.  Si  le  prix  est  1  pour  cent  par  mois,  1  renverse 
indique  le  temps  quand  la  piastre  fait  un  cent,  et 
le  chiffre  change  deux  places  a  la  gauche  montre 
Tintaret  dans  tout  cas. 


Ainsi: 


$  2 

5 

6 

7.35 

8 

6 

1.50 

9, 

7 

4 

6.75 

11, 

4 

6 

3.25 

22, 

5 

3 

8.40 

1, 

0 

0 

0.50 

78  HENDERSON'S 

METHODE  D'EGALISEB  DES  NOMBKES  PAR 
LEUK  COMPLEMENT  ET  SUPPLEMENT. 

Le  complement  d'un  nombre  est  la  difference 
cTun  nombre  et  d'un  autre  nombre  particulier 
avaiit  lui.  Le  supplement  d'un  DS'inbre  est  la  dif- 
f^rence  d'un  nombre  et  d'un  autre  nombre  apres 
lui.  (99)-=9801.  Prenons  le  complement  de  99, 
nous  1'appellerons  centieme,  et  nous  ajouterons  le 
complement  pour  le  rendre  egale, 

EXPLICATION. — Laissons  que  N  e'gale  99,  et  C  egal 
1.  Alors  N  plus  0=100.  N— 0=98.  Multiplions 
les  deux  equivalents  ensemble,  nous  avons  N2 — 
02=9800.  Ajoutons  C2  aux  uns  et  aux  autres 
nombres,  et  nous  aurons  N2— 9801,  le  nombre 
egal  de  99. 

IRE  BEGLE.  —  Quand  le  nombre  est  plus  haut 
que  la  base,  nous  ajoutons  le  supplement,  nous 
1'appellerons  centiemes,  et  nous  ajouterons  Tegal 
du  supplement,  nous  1'appellerons  centieme,  parce- 
que  le  nombre  est  augmente  par  le  supplement 
quand  il  est  multiplie  par  100;  en  ce  cas-ci,  quand 
le  nombre  est  moindre  que  la  base,  nous  sous- 
trairons  le  complement. 

SMS  EEGLE. — Le  produit  de  Tun  ou  de  Tautre  de 
deux  nombres  est  1'dgal  de  leur  valeur  diminue' 
par  F6gal  de  la  moitie  de  leur  difference. 


LIGHTNING  CALCULATOR.  79 

POUR  TROUVER  LA  VALEUR  DE  LA  MON- 
NAIE  COURANTE  QUAND  LE  PRIX  DE 
L'OR  EST  FAIT. 

Quand  Tor  est  a  111-g,  qu'  est  la  valeur  de  $1.00 
enmonnaiecourante?  Nous  prenons  109,  le  nornbre 
de  cent  dans  la  piastre,  cornme  le  numerateur,  et 
la  valeur  de  For  comnie  le  clenominateur.  Noun 
simplifions  la  fraction  en  multipliant  le  numerateur 
et  denominateur  par  9,  et  nous  aurons  le  -f^  d'une 
piastre,  ou  90  cents,  la  valeur  de  la  monnaie  cou- 
rante. 

Quand  la  monnaie  -courante  vaut  75  cents,  quel 
est  la  valeur  de  Tor?  \ff=$,  f  X  1?°  cents  egale 


REGLE.  —  Nous  prenons  100,  le  nombre  de 
cent  dans  une  piastre,  pour  le  numdrateur,  et  la 
valeur  de  For  ou  de  Targent  courant,  quelque  soit 
la  cause,  pour  le  denominateur.  Nous  simplifions 
ia  fraction  en  ajoutant  un  zero  au  numerateur  et  en 
divisant  par  le  denominateur, 

NOUVELLE  METHODS  DECIMAL  DE  CAL- 
CULER  L'INTfiRET  ET  DE  L'EXPLI- 
QUER. 

REGLE.  —  Revers^  le  prix,  ajoutez  un  zero,  et  mettez 
devant  le  point.  Immediatement  au-dessus  de  ces 
caracteres,  placez  le  montant  sur  lequel  I'interSt  es^ 
demande,  le  centieme  etant  toujours  dans  la  co- 
lonne  du  prix. 

RAISONS.  —  Posez  le  prix  pour  trouver  la  regie. 


\  a  R  A  t .  t 

OF  THE 

TT  "NT  T  ^r  T1 -R  QTT1^ 


80  LIGHTNING   CALCULATOR. 

Quand  la  piastre  fait  uncent.  Ajoutez  un  zero,  vous 
trouverez  la  regie,  quand  une  piastre  fait  dix  cents. 
Mettez  devant  le  point,  vous  trouverez  la  regie, 
Quand  une  piastre  fait  un  mille.  Quand  une  piastre 
fait  un  cent,  vous  rechangez  le  point  deux  places  a 
la  gauche:  parce^ue  un  centieme  du  principale 
egale  Finterefc.  Quand  une  piastre  fait  dix  cents, 
rechangez  le  point  une  place  a  la  gauche,  parceque 
le  dixieme  du  principal  est  Tinteret.  Le  prix 
reiiverse  invariablement  represente  le  temps  quand 
une  piastre  gagne  un  cent,  ou  cent  piastres  une 
piastre.  Ainsi,  pour  trouver  quel  que  soit  le  nom- 
bre  donne  de  centiemes,  placez  le  immediatement 
dessous  le  prix  renverse  et  vous  aurez  la  reponse 
en  piastres  et  declines. 


J..  A.  HETOEBSCOT,  A.  M.7 

Phrenologist  and  Phreno-Magnetic  Healer. 


CERTIFICATES  OF  APPROVAL. 


I  have  examined  the  new  methods  of  calcula- 
tion by  Prof.  J.  A.  Henderson,  they  are  inval- 
uable to  business  men,  and  will  prove  a  light 
in  science  to  all  coming  generations. 

A.  J.  WAENEB, 
Pres.  Elmira  Commercial  College. 


Henderson's  methods  are  the  finest  known  for 
lightning  multiplication. 

.Prof.  D.  K.  FOKD, 

Female  College,  Elmira. 


I  have  examined  Prof.  J.  A.  Henderson's  new 
methods  of  calculation;  they  are  remarkable  for 
originality  and  of  great  practical  value.  His 
methods  of  calculating  interest  are  peculiarly 
clear  and  comprehensive  in  their  adaptation  to 
all  possible  cases. 

Rev.  Dr.  O.  P.  FITZGEBALD, 

Ex.  State  Superintendent,  Col. 


CERTIFICATES  OF  APPROVAL.- 

Mr.  J.  A.  Henderson  has  taught  mathematics 
in  Delhi  Academy  for  a  year.     We  consider  him 
an  excellent  mathematical  teacher. 
J.  L.  SAWYER, 

Principal  of  Delhi^.Academy . 
Delhi,  Oct.  1862. 

P.  S.  J.  A.  H.,  taught  analytical  Trignome- 
try,  University  Algebra,  Intellectual  Arithmetic 
and  English  Grammar  in  Delhi  Academy,  New 
York. 


John  Alexander  Henderson,  A.  M.,  attended 
Union  College  and  graduated  with  me  in  class 
"64."     He  is  an  excellent  scholar — among  the 
first — and  his  character  is  above  reproach. 
ELISHA  CURTIS,  A.  M., 

Principal  of  Sodus  Academy. 


I  have  known  Prof.  J.  A.  Henderson  from 
earliest  boyhood;  his  character  has  always  been 
beyond  reproach.  As  a  mathematician  he  has 
scarcely  an  equal ;  as  a  teacher  he  has  been  emi- 
nently successful;  as  a  phrenologist,  he  is  con- 
sidered by  many  not  a  whit  behind  Fowler  & 
Wells.  New  York. 

Rev  A.  G.  KING, 

of  U.  P.  Church,  N.  Y.,  1869. 


EXTBACT  FBOM  J.  A.  HENDERSON'S  PHBENO- 
MEDICAL  CHABT. 

Diet,  exercise,  rest,  light,  good  water  and  pure 
air  develop  the  mental,  motive  and  vital  forces. 
Young  maiden  and  young  man,  make  these  your 
physicians,  for  they  insure  health,  success  in 
business,  give  you  the  key  to  philosophy  and 
are  the  handmaids  of  Christianity.  Add  in- 
telligence and  contentment,  the  two  great  pillars 
of  felicity,  and  you  do  much  to  sustain  the  moral 
government  of  the  domestic  circle,  the  moral 
government  of  the  human  family,  and  the  moral 
government  of  the  Creator. 

Strong  faith  makes  a  stout  heart;  active  hope, 
a  healthy  liver;  rounded  up  veneration,  excellent 
functions  of  digestion;  large  firmness,  strong 
vertebrae;  fine  conscientious  gives  not  only  a 
pure  mind,  but  healthy  kidneys  eliminating  all 
surplus  secretion  from  the  brain  and  body; 
large  combativeness  develops  a  fine  osseous 
force;  destructiveness  an  excellent  muscular 
force.  The  mind  is  the  root,  the  body  the 
trunk,  and  the  sciences  the  branches.  The  seat 
of  the  mind  is  the  brain  and  nerves.  The 
organs,  forty  in  number,  are  the  instruments 
used  in  framing  constitutions  and  building  up 
science.  They  are  also  the  instruments  used  in 


CERTIFICATES  OF  APPKOVAL. 

building  up  the  constitution  of  the  body.  There- 
fore beware  that  you  build  in  no  error,  for  dis- 
ease will  surely  follow,  which  is  the  result  of 
insulted  law.  It  is  this  law  insulted  that  binds 
the  body  with  disease;  break  the  bands  and 
your  blood  will  flow  like  wine,  and  disease  dis- 
appear like  mist  before  the  morning  sun.  Mind  is 
light.  Hail !  holy  light  which  lighteth  everyone, 
in  tJiee  is  the  life  of  the  blood,  flesh  and  spirit, 
from  thee  the  face  gets  its  form  and  beauty,  the 
eye  its  light,  the  tongue  the  word,  the  muscle 
its  action,  and  every  function  its  health  and  de- 
velopment. Hence  keep  all  the  organs  of  the 
mind  and  functions  of  the  body  well  ventilated 
by  being  correct  in  diet,  exercise,  rest,  light, 
good  water  and  pure  air,  and  the  result  is  sound 
flesh  and  pure  blood. 

The  child  is  the  zero  power  of  its  parents, 
that  is  a  unit  of  the  parents,  plus  or  minus  the 
surrounding  influences.  A  well  balanced  child 
has  its  eyes  in  the  center  of  its  head,  that  is  the 
same  distance  from  the  point  of  the  chin  to  the 
optics,  as  from  the  optics  to  the  upper  part  of 
the  organ  of  benevolence. 

A  wise  man's  eyes  are  in  the  center  of  the 
head.  The  face  is  an  index  of  the  strength  of 
the  will  of  the  flesh;  the  brain  an  index  of  the 
strength  of  the  will  of  the  mind.  Hence  when 
Nature  establishes  an  equilibrium  between  the 
two  forces,  the  master  and  the  servant,  the  will 
of  the  mind  and  the  will  of  the  flesh,  wre  have 
harmony  of  character,  prudence,  wisdom  and 
proficiency.  The  history  of  aM  nations  illus- 
trates the  truth  of  the  above  propositions  very 
clearly. 

J.  A.  Henderson  is  preparing  a  lightning 
.  of  discerning  character. 


METODO  IKSTANTAMO 

DE  CALCULAB 

Intelectnal  y  Practical 


PROFESOR  J.  A.  HENDERSON, 

SRADUADO  EN  EL  COLEGIO  UNION, 

Y  AUTOR  DEL  CALCULADOR,  LIBRO  DE  "CUADRO   ILOS- 
TRANDO  RAICES,  Y  CARTA  PRENA-MEDICAL. 


Examinacions  tocante  at  salted  negocios  y  otras 
cosas,  5&2  Calle  de  Market,  Cuatro  ^K°.  16. 


SAN  FRANCISCO: 

IMPRENTA  C08MOPOLITANA,  CALLE  DE  CLAY,  NO. -505. 

1872. 


K 


EL  CALCUIADOK  INSTANTAflEO. 


El  Alfabeto  numerico,  segun  se~escribe  y  se 
lee: 


r     §• 


I'    I-    I    I    I 

O 


p. 

B      PI 

O  £ 

W         O 
' 


El  segundo 
es   dos   veces 
el  primero, 
__   l_l_JLl_   l_.L_.Ll__Lel  tercero  tres 
^7890.  veces    el   pri- 
l'   l'   l'   l'   l'   l'   l'   l'   i'   i   mero,    etc., 
hasta  el  ultimo. 

KEGLA  DE  ESTTEEES. 

REGLA. — El  reciproco  de  la  tasa  6  la  tasa  in- 
vertida,  indica  el  tiempo  cuando  en  todos  casos. 
Se  puede  mudar  el  tiempo  decimal  dos  lugares 
hacia  la  izquierda;  diez  veces  aquel  tiempo  uii 
lugar  hacia  la  izquierda  y  un  desimo  del  mismo 
tiempo  tres  lugares  hacia  la  izquierda.  Aumen- 
tase  6  disminuyase  el  resultado  para  que  coii- 
cuerde  con  el  tiempo  dado. 


90 


EL  CALCULADOB  INSTANTANEO 


Tasa  por  ano.  ...... 

9 

"    invertido.......  .    . 

1 

0 

Con  diez  desimos*,,. 

9 

$1 

9 

7 

dias. 

40 
dias 

400 
dias 

|  por  mes...  ......... 

3 

% 

8 

0 

3 

EJEMPLO-^  .  „  $15 

4 

9 

8 
dias. 

80 
dias 

• 

800 
dias 

"o 

i 

Unidades 

<o 

ft 

6,21 


6.25 


A  un  centavo  por  mes  6  1%.     Por  ejemplo : 


M             M             M 

p         u         a 

r 

f" 

?• 

P- 

p. 

CD 

o 

CO 

1 

s 

i 

$27 

~~5~ 

IT 

7.35 

8 

6 

1.50 

9, 

7 

4 

6.75 

11, 

4 

6 

3.35 

22, 

5 

3 

8.40 

1, 

0 

0 

0.50 

DE   HENDERSON.  91 

Para  obtener  resultados  por  cualquier  otro 
tiempo  dado,  se  aumenta  6  divide  segun  sea  el 
caso. 

N.  B. — Se  puede  aplicar  el  mismo  metodo  a* 
cualquier  ejemplo  6  tasa  de  inter es. 


METODO    DE     CUADKAR    NUMEBOS    FOB    SFS    SUPLI- 
MENTO   Y     COMPLIMENTO. 

El  complimento  de  un  numero  es  la  diferencia 
entre  aquel  numero  y  otro  numero  especial  ma- 
yor que  aquel.  El  suplimento  de  un  numero  es 
la  diferencia  entre  aquel  numero  y  otro  numero 
especial  menor  que  aquel. 

992=9,801.  Eestese  el  complimento  de  99, 
llamase  cientos  y  an^dasele  el  cuadrado  del 
complimento. 

ESPLICACION. 

Supongamos  por  ejemplo,  que  la  n  iguale  & 
(99)  y  la  c  iguale  a  1.  Entouces  n  mas  c=100. 
n — c=98.  Multiplicand  las  dos  equaciones, 
tendremos  n2 — c2=9,800.  Anadanse  c2  &  am- 
bas  cantidades  de  la  equacion  y  tendremos  n2 
=9,801,  el  cuadrado  de  99. 

REGLA. — Cuando  mayor  que  la  base,  el  cual 
es  ciento  anadase  el  suplimento,  IMmesele  cien- 
tos y  anadasele  el  cuadrado  del  suplemento  llam- 
esele  cientos,  por  que  el  numero  cuando  aumen- 
tado  por  el  suplemento  es  en  este  caso  multipli- 


92       EL  OALCULADOR  INSTANTANEO 

cado  por  ciento,    cuando  es  menor  r^stese  el 
complimento. 

(19)  1  361.  El  complimento  es  uno  1,  de  19 
restamos  1  quedan  18,  18  multiplicado  por  20 
es  igual  a  360  anadase  el  cuadrado  de  1  y  tendre- 
mos  el  cuadrado  de  19.  (49)  1 2,401.  El  com- 
plimento es  1,  de  49  restamos  1  quedan  48  llam- 
esele  quincuajesimo,  anadasele  el  cuadrado  de 
1,  y  tenernos  2,401,  por  resultado. 

EEGLA. — El  producto  de  dos  numeros  cualqui- 
era  es  el  cuadrado  de  su  media  suma  dismi- 
nuido  por  el  cuadrado  de  la  mitad  de  su  dife- 
rencia. 

EL  MAS  COMUN  FACTOR  O  DIVISOR. 

EEGLA. — Separese  los  numeros  en  sus  factores 
primitives .  El  producto  de  todos  los  factores 
que  son  comunes  sera*  el  mas  comun  divisor. 

£  Cual  es  el  mas  comun  divisor  de  21  y  77  ? 
Separando  los  numeros  en  sus  factores  primiti- 
vos,  tenemos  21=7x3,  77=7x11,  de  consigui- 
ente,  7  es  el  mas  comun  factor,  6  divisor  de  los 
doB  numeros. 

BEGLA  PAEA  ANADIR  Y  RESTAR  QUEBEADO9. 

Eeduzcanse  primeramente  todos  los  quebra- 
dos  a  un  mismo  denominador,  anadanse  los  nu- 
meradores  y  pongase  la  suma  sobre  el  denomi- 
nador comun.  Para  restar  escribase  la  difer- 


DE   HENDERSON. 

encia  de  los  numeradores  sobre  el  comun  deno- 
minador. 
EJEMPLO.  —  <<  Cuanto  >es  la  suma  de^='l  J=^ 


EJEMPLO.  —  De  f  restase  J=£ 

DIVISION  DE  QUEBEADOS. 

KEGLA  —  Reduzcanse  los  niimeros  fraccionarios 
a*  la  forma  de  quebrados  impropios;  multipli- 
quese  el  dividiendo  por  el  divisor  invertido  6 
multiplfquese  tanto  el  numerador  y  denomina- 
dor  por  el  mas  comma  multiplicador  de  los  de- 
nominadores  de  los  partes  fraccionales. 

Dividase  5J  por  2J-,  multiplfquese  tanto  el 
numerador  y  denominador  por  6,  el  mas  comun 
de  2  y  3. 


PABA  HALLAB  EL  VALOB  DE  MONEDA  O  PAPEL  MONEDA 
CUANDO  SE  CONOCE  EL  PBEGIO  DEL  OBO. 

EEGLA.---Tomamos  100,  el  numero  de  centa- 
vos  que  contiene  el  peso  fuerte  para  numerador 
y  el  valor  del  oro,  6  papel  moneda,  segun  sea 
el  caso,  'para  denominador.  Simplifiquese  el 
quebrado  anadiendo  ceros,  al  numerador  y 
dividiendo  por  el  denominador. 

Cuando  el  oro  esta*  al  valor  de  109J. 

<:  Cuanto  es  el  valor  de  $1  .  00  moneda  papel  ? 

EJEMPLO.-H0  =^-l^-$91  ** 
—      ~ 


982  —  491~          401. 


94        EL  CALCULADOK  INSTANTANEQ 

Cuando  el  papel  moneda  esta  al  valor  de  75 
centavos,  £  cuanto  es  el  valor  del  oro  ? 

EJEMPLO — ^=t  |  de  100  centavos  igual  a  $1. 
33J.  

METODO  DE  EXTRAER  LA  KAlZ  CUADRADA. 

EEGLA. — Dividase  cualquier  niimero  por  el 
cuadrado  de  tres,  estraigase  la  rafz  cuadrada 
del  cociente  y  tenemos  un  tercio  de  la  raiz  del 
niimero.  Dividase  cualquier  niimero  por  el 
cuadrado  de  cuatro,  estraigase  la  raiz  cuadrada 
del  cociente  y  tenemos  un  cuarto  de  la  raiz 
cuadrada  del  niimero,  etc, 

32      1 

100    15625(100+20+5 
Primer  divisor  probante  200     10000 


Segundo 


verdadero  220 

5625 

probante  240 

4400 

verdadero  235 

1225 

1225; 

RAIZ  CUBICA. 


Anadase  a  cada  verdadero  divisor,  segun  se 
presenten  dos  veces  la  supe^cie  de  un  lado 
del  pequeno  tubo,  y  uno  a  cada  uno  de  los  tres 
rectangolos  para  el  divisor  probante,  porque  eso 
hara  los  tres  lados  del  cubico  entero. 


BE  HENDERSON.  95 

REGLA. — Dividase  cnalquier  mimero  por  el  cu- 
bico  de  2,  extraigase  la  raiz  ciibica  del  cociente, 
y  tenemos  la  raiz  cubica  del  mimero.  Dividase 
cnalquier  mimero  por  el  ciibico  de  3,  6  27,  ex- 
traigase la  raiz  cubica  del  cociente  y  tenemos  un 
tercio  de  la  raiz  del  mimero.  Dividase  cual- 
quier  mimero  por  el  ciibico  de  4,  6  64,  extrai- 
gase la  raiz  cubica  del  cociente  y  tenemos  un 
cuarto  de  la  raiz  del  mimero.  Dividase  cual- 
quier  mimero  por  el  cubo  de  5,  6  125  extraigase 
la  raiz  cubica  del  cociente  y  tenemos  un  quinto 
de  la  raiz  del  mimero. 

1728)  110592  (64 
'   10368 

6912 
6912 


PARA  ENCONTRAR  EL  NUMERO  DE  PIES  €UBICOS~QUE 
CONTIENE   EL  BUSHEL. 

EEGLA. — Mudase  el  punto  decimal  un  lugar 
hacia  la  izquierda  y  multipliquese  el  cociente 
por  8. 

EJEMPLO.— Dentro  de  un  vacillo  de  800.9  pies 
ciibicos,  raudase  el  punto  un  lugar  hacia  la  iz- 
quierda, y  tenemos  80.09;  multipliquese  por  8 
y  tenemos  640 . 72  el  numero  de  bushels. 

N.  B. — Anadase  uno  por  cada  tres  cientos . 


96  EL   CALCULATOR  INSTANTANEO 

EJEBCICIO   MENTAL. 

El  mimero  medio  de  dos  numeros  es  la  mitad 
de  sus  sumas  6  el  numero  igualmente  distante 
de  los  dos  rnimeros.  El  producto  de  dos  nii- 
meros,  es  el  cuadrado  del  numero  medio  dismi- 
nuido  por  el  cuadrado  de  la  mitad  de  su  dife- 
rencia. 

PEOBLEMA. — 19  por. 21,  18  por  22,  etc.,  hasta 
15.  Asi:  el  numero  medio  es  20,  el  cuadrado 
de  20  es  400,  400— el  cuadrado  de  1  es  399;  el 
producto,  18  por  22  es  el  cuadrado  de  200,  400 
el  cuadrado  de  2,  4,  396.  17  por  23  es  391,  16 
por  24,  384;  15  por  25,  375. 


El  complimento  de  un  numero  es  la  diferencia 
de  aquel  numero  y  algun  otro  numero  especial 
mayor.  El  suplemento  de  un  numeroes  la  dife- 
rencia de  aquel  mimero  y  algun  otro  numero 
especial  menor. 

Asi  el  complimento  de  99  es  la  diferencia  de 
99  y  100,  el  cual  es  uno.  El  suplemento  de  101 
es  la  diferencia  de  101  y  100,  el  cual  es  1. 


ticue 


X  31.  i^enbcrson, 

un&  urn  fcicfel&e  leidjt  Bcgmflufj  ju  tuadjen. 

Die  $?a§regel  betfelben  finb  fotgenber  2Irt  : 
1.  Das  numerirte  Sllpljatet  femten  ju  lernen/ 
wie  folgt  : 

1234567890 


2*  X)a^    jiipulirte   Sntereffe   na^ 
ximjubre^en,  mt  folgt  : 

3  pr.  fit.  per  SHonat  ober  5;  umgebre^t  ~  5Wonat 
f  10  £age. 

3.  giir  ba^  jel)nfad)e  eine  9luK  .  ober  Zero  ju 
anenren,  mie  folgt  : 

i  9Jionatxbet  10  ober  f  S)?onat  ^  100  Sage. 

4,  giit  beit  jefynten  Sfytil  eine  S^utt  (Zero)  ju 
prafixiren,  wic  folgt  : 

^bet  10  ober     9Honat  -  1  Sa,— 


Die  ©riinbe  ftnb  folgenbe  : 

Umbrefjen  ber  Sntereffen  bcmonjitht 
immct  bte  B^^,  tDenn  em  Staler  ($  1.00)  einen 
Sent  (1  ct.)  ma$t. 


98 


anenren  einer  9Zutt  (0)  aerjetynfadjt  bic 
Beit,  gteicfy  bieS 

1  Sent  ju  ]  0  gents. 

£>a«  prajmren  eines  ^unites  ober  9lutt  (0) 
aerliirjt  bte  gtit  je^nfad^f  ba^er  an^  bet  Sent 
ate  etne  $tttte  ober  ^  Sent  gere^ttcttt>etbenniu§. 


3  pr.  <£t  pet  SKonat  [tftx] 

3 

3 

a$  bent  mtfjabet  --• 

. 

1 

0 

3 

umgebre^t  ma^t  e*  y3  per  9ftenat  ober  10  5Taije. 

*M 

0 

1 

1 

1-4 

o 

| 

$  15 

0 

0 

o.oo 

9 

0 

0 

0.00 

1 

2 

0 

0.75 

1 

3 

6 

0.25 

(£r  ft  ar  un  g  . 

3ebermann,  ber  baljer  bte  oben  gcliefcrteu 
regetn  griinblt^  geternt  ^at,  wirb  leictyt  etn^ 
,  mit  n>eld)er  Sei^tigleit  btefe  9J?etl;obe  atte 


99 


JRecfottimgcn   lofH  inbem    fte    bas 
Scfyanerige  burd)  Decimals  angreift 

£at  man  baljer  ausgefunben  menu  $  1. 00  ei* 
uen  Sent  macfyt,  fo  nnrb  Sebermann  flat  einfe* 
1)011,  ba£  er  nur  bie  ftoftl  bet  Better  aU  Sent^ 
ju  betrad;ten  fyat,  um  bie  Sofung  berfelien  ju 
ftnben, 

©ottte  man  nun  iDiinfi^en  au^juftnben  bie 
Sntereffen  i^on  enter  ©utnmc  ju  3  per  Sent  per 
Sftonat  fur  1 33  Sage,  fo  gtbt  bte  $enntnt§  btefer 
9)Jet|)obe  efcenfalt^  etne  letcfyte  Sofung. 

[3nm  93eifpiel]  Die  Sntereffen  »on  $  1,00 
ju  3  per  Sent  per  Sftonat : 

3  per  St.  per  SRonat  ift  ^  J  SKonat  *  10  Sage 
(unb  ntacfyt  1  Sent  in  10  Sagen)  fotgtt^  fur 
1 33  Sage 

fiir  100  Sage  t>a3  setynfadje  i\  10  Sagen  ot).  iSt^.  =  10 
„  30  „  ba^3x  „  „  10  „  „  L  „  -  3 
„  3  „  3  mal  b.  loten  Jfeeil »,  10 1.  n  \  „  *  0,3 

Kent*  13,3 


HENDERSON'S  CHART, 

For  Computing  Time  and  Interest,  Squaring  and  Mul- 
tiplying Numbers,  Dividing  Fractions,  etc.  Copyrighted 
January  22,  1872.  Price,  $1.00. 


A  BOOK  OF  BLOCKS, 

ILLUSTRATING  ROOTS. 

For  Schools,  Academies   and  Colleges. 

Copyrighted  May  11,  1872.     Price,  §5.00. 


HENDERSON'S  INTELLECTUAL   AND 

PRACTICAL  LIGHTNING 

CALCULATOR. 

Copyrighted  October  24,  1872.     Price,  $1.00. 


NEW    DECIMAL     METHOD    OF    COM- 
PUTING AND  IMPARTING 
INTEREST, 

ILLUSTRATED  BY  AN  ENGRAVING. 

Copyrighted  November  20,  1872.     Price,  25  cents. 


AUTHOR'S   NOTICE. 

Instruction  gratis  to  those  who  purchase  the  above  of 
the  Author  at  his  office,  No.  542  Market  street,  San 
Francisco.  Office  Hours — From  10  to  11  A.M.,  and  from 
4  to  5  P.M. 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


MAR  12 


- 


LD  21-100ra-7,'39(4<" 


